C o d e F o r c e s 264 B CodeForces 264B CodeForces264B G o o d S e q u e n c e s Good Sequences GoodSequences

思路：
质因数分解后，求当前数的质因数种出现的最大次数，然后当前数的所有质因数赋值当前的最大，进行 d p dp dp即可，时间复杂度 O ( n l o g n ) O(nlogn) O(nlogn)

/** @Author: vain* @Date: 2020* @LastEditTime: 2020-10-06 13:40:49* @LastEditors: sueRimn* @Description: 学不会 dp 的 fw* @FilePath: \main\demo.cpp*/
//#include <bits/stdc++.h>
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <ctime>
#include <cstring>
#include <cstdlib>
#include <math.h>
#include <bitset>
//#include <unordered_map>
using namespace std;
typedef long long ll;
#define ll long long
//typedef unsigned long long uint;
const int N = 5000 + 20;
const int maxn = 1e5 + 20;
const int mod = 998244353;
ll in[maxn], cnt, head[maxn];//typedef pair<ll, ll> p;
//priority_queue<p, vector<p>, greater<p>> q;
//int sum[maxn];
double Max(double a, double b) { return a > b ? a : b; }
int min(int a, int b) { return a < b ? a : b; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a * b / gcd(a, b); }
void swap(int &x, int &y) { x ^= y, y ^= x, x ^= y; }
int lowbit(int x) { return (x) & (-x); }
map<int, int> mp;
ll ksm(ll a, ll b)
{ll res = 1;for (; b;){if (b & 1)res = (res * a) % mod;b >>= 1, a = a * a % mod;}return res;
}
int read(int &v)
{int k = 1;v = 0;int c = getchar();while (c < '0' || c > '9'){if (c == '-')k = 0;c = getchar();}while (c >= '0' && c <= '9')v = (v << 3) + (v << 1) + (c - 48), c = getchar();if (k == 0)v = -v;return c;
}
vector<int> f;
int a[maxn];
int dp[maxn];
int solve(int x)
{int k = x, res = 0;for (int i = 2; i * i <= x; i++){bool fla = true;while (x % i == 0){if (fla)dp[i]++, fla = false;x /= i;res = max(dp[i], res);}}if (x > 1)dp[x]++, res = max(dp[x], res);x = k;for (int i = 2; i * i <= x; i++){bool fla = true;while (x % i == 0){x /= i;if (fla)fla = false, dp[i] = res;}}if (x > 1)dp[x] = res;return res;
}
int main()
{int n;read(n);for (int i = 1; i <= n; i++){read(a[i]);}int ans = 1;for (int j = 1; j <= n; j++){ans = max(ans, solve(a[j]));}//cout << endl;printf("%d\n", ans);
}

C o d e F o r c e s 246 D CodeForces 246D CodeForces246D C o l o r f u l G r a p h Colorful Graph ColorfulGraph

思路：
分组，染色，求贡献

/** @Author: vain* @Date: 2020* @LastEditTime: 2020-10-06 19:56:23* @LastEditors: sueRimn* @Description: 学不会 dp 的 fw* @FilePath: \main\demo.cpp*/
#include <bits/stdc++.h>
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <ctime>
#include <cstring>
#include <cstdlib>
#include <math.h>
#include <bitset>
//#include <unordered_map>
using namespace std;
typedef long long ll;
#define ll long long
//typedef unsigned long long uint;
const int N = 5000 + 20;
const int maxn = 1e5 + 20;
const int mod = 998244353;
int cnt, head[maxn], pos[maxn];//typedef pair<ll, ll> p;
//priority_queue<p, vector<p>, greater<p>> q;
//int sum[maxn];
double Max(double a, double b) { return a > b ? a : b; }
int min(int a, int b) { return a < b ? a : b; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a * b / gcd(a, b); }
void swap(int &x, int &y) { x ^= y, y ^= x, x ^= y; }
int lowbit(int x) { return (x) & (-x); }
map<int, int> vis;
//unordered_map<int,int>vis;
ll ksm(ll a, ll b)
{ll res = 1;for (; b;){if (b & 1)res = (res * a) % mod;b >>= 1, a = a * a % mod;}return res;
}
int read(int &v)
{int k = 1;v = 0;int c = getchar();while (c < '0' || c > '9'){if (c == '-')k = 0;c = getchar();}while (c >= '0' && c <= '9')v = (v << 3) + (v << 1) + (c - 48), c = getchar();if (k == 0)v = -v;return c;
}
struct node
{int v, nex;
} edge[maxn << 1];
void add(int u, int v)
{edge[cnt].v = v, edge[cnt].nex = head[u];head[u] = cnt++;
}
vector<int> vec[maxn], f;
int main()
{int n, m, color = 100000;read(n), read(m);for (int i = 1; i <= n; i++){head[i] = -1;read(pos[i]);f.push_back(pos[i]);vec[pos[i]].push_back(i);if (pos[i] < color)color = pos[i];}sort(f.begin(), f.end());f.erase(unique(f.begin(), f.end()), f.end());for (int i = 1; i <= m; i++){int u, v;read(u), read(v);add(u, v), add(v, u);}int ans = 0, k = f.size();for (int i = 0; i < k; i++){int maxx = 0;vis.clear();for (int j = 0; j < vec[f[i]].size(); j++){int p = vec[f[i]][j];for (int u = head[p]; ~u; u = edge[u].nex){int v = edge[u].v;if ((pos[v] != pos[p]) && (!vis[pos[v]]))maxx++, vis[pos[v]] = 1;}if (maxx > ans)color = pos[p], ans = maxx;else if (maxx == ans && pos[p] < color)color = pos[p];}}printf("%d\n", color);
}

C o d e F o r c e s 369 C CodeForces 369C CodeForces369C V a l e r a a n d E l e c t i o n s Valera and Elections ValeraandElections

思路：
枚举叶节点，边权转化，找最先出现的坏点
数据水了，被俩队友hack了，最坏时间复杂度是 O ( N 2 ) O(N^2) O(N2)，但记录一下

/** @Author: vain* @Date: 2020* @LastEditTime: 2020-10-11 12:08:33* @LastEditors: sueRimn* @Description: 学不会 dp 的 fw* @FilePath: \main\demo.cpp*/
#include <bits/stdc++.h>
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <ctime>
#include <cstring>
#include <cstdlib>
#include <math.h>
#include <bitset>
//#include <unordered_map>
using namespace std;
typedef long long ll;
#define ll long long
//typedef unsigned long long uint;
const int N = 5000 + 20;
const int maxn = 1e5 + 20;
const int mod = 998244353;
//typedef pair<ll, ll> p;
//priority_queue<p, vector<p>, greater<p>> q;
//int sum[maxn];
double Max(double a, double b) { return a > b ? a : b; }
int min(int a, int b) { return a < b ? a : b; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a * b / gcd(a, b); }
void swap(int &x, int &y) { x ^= y, y ^= x, x ^= y; }
int lowbit(int x) { return (x) & (-x); }
//map<int, int> vis;
ll ksm(ll a, ll b)
{ll res = 1;while (b){if (b & 1)res = (res * a) % mod;b >>= 1, a = a * a % mod;}return res % mod;
}
int read(int &v)
{int k = 1;v = 0;int c = getchar();while (c < '0' || c > '9'){if (c == '-')k = 0;c = getchar();}while (c >= '0' && c <= '9')v = (v << 3) + (v << 1) + (c - 48), c = getchar();if (k == 0)v = -v;return c;
}
struct node
{int v, pos, nex, w;
} edge[maxn << 1];
int cnt, head[maxn], pre[maxn], vis[maxn];
int dwn[maxn], in[maxn], color[maxn];
vector<int> f;
void add(int u, int v, int w)
{edge[cnt].v = v, edge[cnt].w = w;edge[cnt].nex = head[u], head[u] = cnt++;
}
void dfs(int u, int fa)
{pre[u] = fa;for (int i = head[u]; ~i; i = edge[i].nex){int v = edge[i].v;if (v != fa){if (edge[i].w == 2)vis[v] = 1;dfs(v, u);}}
}
int main()
{int n;read(n);for (int i = 1; i <= n; i++)head[i] = -1;for (int i = 1; i < n; i++){int u, v, w;read(u), read(v), read(w);in[v]++, in[u]++;add(u, v, w), add(v, u, w);}dfs(1, 0);for (int i = 2; i <= n; i++){if (in[i] == 1){f.push_back(i);}}int m = f.size();for (int i = 0; i < m; i++){int k = f[i];bool fla = false;while (k != 1){if (color[k])break;if (fla)color[k] = 1;if (vis[k]){dwn[k]=1;fla = true;}k = pre[k];}}f.clear();int ans = 0;for (int i = 1; i <= n; i++){if (dwn[i] == 1 && (!color[i]))ans++, f.push_back(i);}printf("%d\n", ans);bool fla=false;for (int i = 0; i < ans; i++){fla=true;printf("%d ", f[i]);}if(fla)puts("");
}

C o d e F o r c e s 815 A CodeForces 815A CodeForces815A K a r e n a n d G a m e Karen and Game KarenandGame
思路：考虑两种状况 ， n > m n>m n>m和 n < = m n<=m n<=m

参考代码

#include<iostream>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<unordered_map>
#include<unordered_set>
#include<set>
#include<cstdlib>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<set>
using namespace std;
typedef long long ll;
const int maxn=2e5+5;
const int N=5e2+5;
int read(int &v)
{int k = 1;v = 0;int c = getchar();while (c < '0' || c > '9'){if (c == '-')k = 0;c = getchar();}while (c >= '0' && c <= '9')v = (v << 3) + (v << 1) + (c - 48), c = getchar();if (k == 0)v = -v;return c;
}
ll ksm(ll a,ll b)
{ll res=1;while(b){if(b&1)res*=a;b>>=1,a=a*a;}return res;
}
int mp[105][105],sum=0;
int ans[3][105];
int dp[105][105];
int dx[105][105];
void solve1(int n,int m)//先算行
{for(int i=1; i<=n; i++){int minx=505;for(int j=1; j<=m; j++)minx=min(minx,mp[i][j]);if(minx)ans[0][i]=minx,sum+=minx;dp[i][1]-=minx,dp[i][m+1]+=minx;for(int j=1; j<=m; j++)mp[i][j]=dp[i][j]+mp[i][j-1];}
}
void solve2(int n,int m)//先算列
{for(int i=1; i<=m; i++){int minx=505;for(int j=1; j<=n; j++)minx=min(minx,mp[j][i]);if(minx)ans[1][i]=minx,sum+=minx;dx[1][i]-=minx,dx[n+1][i]+=minx;for(int j=1; j<=n; j++)mp[j][i]=dx[j][i]+mp[j-1][i];}
}
void init1(int n,int m)
{for(int i=1; i<=m; i++){for(int j=1; j<=n; j++){dx[j][i]=mp[j][i]-mp[j-1][i];}}
}
void init2(int n,int m)
{for(int i=1; i<=n; i++){for(int j=1; j<=m; j++){dp[i][j]=mp[i][j]-mp[i][j-1];}}
}
int main()
{int n,m;read(n),read(m);for(int i=1; i<=n; i++){for(int j=1; j<=m; j++){read(mp[i][j]);}}if(n>m){init1(n,m);solve2(n,m);init2(n,m);solve1(n,m);}else{init2(n,m);solve1(n,m);init1(n,m);solve2(n,m);}for(int i=1; i<=n; i++){for(int j=1; j<=m; j++){if(mp[i][j]){puts("-1");return 0;}}}printf("%d\n",sum);for(int i=1; i<=n; i++){while(ans[0][i]--){printf("row %d\n",i);}}for(int i=1; i<=m; i++){while(ans[1][i]--){printf("col %d\n",i);}}
}

C o d e F o r c e s 711 C CodeForces 711C CodeForces711C C o l o r i n g T r e e s Coloring Trees ColoringTrees
思路
暴力转移dp，用一个三维的 d p dp dp，记录第 i i i颗树，使用第 j j j种颜色，分了 k k k组的最小贡献，时间复杂度 O ( N 3 ) O(N^3) O(N3)
参考代码

#include<iostream>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<unordered_map>
#include<unordered_set>
#include<set>
#include<cstdlib>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<set>
using namespace std;
typedef long long ll;
const int maxn=2e5+5;
const int N=5e2+5;
int read(int &v)
{int k = 1;v = 0;int c = getchar();while (c < '0' || c > '9'){if (c == '-')k = 0;c = getchar();}while (c >= '0' && c <= '9')v = (v << 3) + (v << 1) + (c - 48), c = getchar();if (k == 0)v = -v;return c;
}
ll ksm(ll a,ll b)
{ll res=1;while(b){if(b&1)res*=a;b>>=1,a=a*a;}return res;
}
ll dp[105][105][105];
int cost[105][105];
int color[105],reg[105][105];
int main()
{int n,m,k;read(n),read(m),read(k);for(int i=1; i<=n; i++)read(color[i]);for(int i=1; i<=n; i++){for(int j=1; j<=m; j++){read(cost[i][j]);}}for(int i=0; i<=n; i++){for(int j=0; j<=m; j++){for(int l=0; l<=k; l++){dp[i][j][l]=100000000000000000;}}}if(!color[1]){for(int i=1; i<=m; i++){dp[1][i][1]=cost[1][i];}}else{dp[1][color[1]][1]=0;}for(int i=2; i<=n; i++){if(!color[i]){for(int j=1; j<=m; j++){for(int a=1; a<=m; a++){for(int b=1; b<=k; b++){if(a!=j)dp[i][j][b]=min(dp[i-1][a][b-1]+cost[i][j],dp[i][j][b]);elsedp[i][j][b]=min(dp[i-1][a][b]+cost[i][j],dp[i][j][b]);}}}}else{int j=color[i];for(int a=1; a<=m; a++){for(int b=1; b<=k; b++){if(color[i]!=a)dp[i][j][b]=min(dp[i-1][a][b-1],dp[i][j][b]);elsedp[i][j][b]=min(dp[i-1][a][b],dp[i][j][b]);}}}}ll minx=100000000000000000;for(int i=1;i<=m;i++){minx=min(dp[n][i][k],minx);}if(minx==100000000000000000ll)puts("-1");elseprintf("%lld\n",minx);
}

C o d e F o r c e s 409 H CodeForces 409H CodeForces409H A + B S t r i k e s B a c k A + B Strikes Back A+BStrikesBack
思路：没有
愚人节的一道题目，然而并没有什么思维上的考察

参考代码

#include<bits/stdc++.h>
using namespace std;
int main()
{int a,b;scanf("%d %d",&a,&b);printf("%d\n",a+b);
}

C o d e F o r c e s 494 A CodeForces 494A CodeForces494A T r e a s u r e Treasure Treasure
思路
模拟栈，先将可以匹配的下标匹配，对剩下的下标进行操作即可
参考代码

/** @Author: vain* @Date: 2020* @LastEditTime: 2020-10-11 09:42:28* @LastEditors: sueRimn* @Description: 学不会 dp 的 fw* @FilePath: \main\demo.cpp*/
#include <bits/stdc++.h>
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <ctime>
#include <cstring>
#include <cstdlib>
#include <math.h>
#include <bitset>
//#include <unordered_map>
using namespace std;
typedef long long ll;
#define ll long long
//typedef unsigned long long uint;
const int N = 5000 + 20;
const int maxn = 1e5 + 20;
const int mod = 998244353;
int cnt, head[maxn], pos[maxn];
//typedef pair<ll, ll> p;
//priority_queue<p, vector<p>, greater<p>> q;
//int sum[maxn];
double Max(double a, double b) { return a > b ? a : b; }
int min(int a, int b) { return a < b ? a : b; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a * b / gcd(a, b); }
void swap(int &x, int &y) { x ^= y, y ^= x, x ^= y; }
int lowbit(int x) { return (x) & (-x); }
//map<int, int> vis;
ll ksm(ll a, ll b)
{ll res = 1;while (b){if (b & 1)res = (res * a) % mod;b >>= 1, a = a * a % mod;}return res % mod;
}
int read(int &v)
{int k = 1;v = 0;int c = getchar();while (c < '0' || c > '9'){if (c == '-')k = 0;c = getchar();}while (c >= '0' && c <= '9')v = (v << 3) + (v << 1) + (c - 48), c = getchar();if (k == 0)v = -v;return c;
}
char s[maxn];
int sta[maxn], sum[maxn], ans[maxn];
int main()
{scanf("%s", s + 1);int n = strlen(s + 1);int top = 0;for (int i = 1; i <= n; i++){if (s[sta[top]] == '(' && s[i] == ')')top--;else if (s[i] != '#')sta[++top] = i;}for (int i = 1; i <= n; i++){if (s[i] == '#')sta[++top] = i;}sort(sta + 1, sta + 1 + top);bool fla = true;for (int i = 1; i <= top; i++){if (s[sta[i]] == ')'){fla = false;break;}if (s[sta[i]] == '(')sum[i] = sum[i - 1] + 1;elsesum[i] = sum[i - 1];}int res = sum[top];int x = 0;for (int i = 1; i <= top; i++){sum[i] -= x;if (s[sta[i]] == '#' && sum[i])ans[sta[i]] = 1, x++, sum[i]--, res--;else if (s[sta[i]] == '#' && (!sum[i])){fla = false;break;}}for (int i = top; i >= 1; i--){if (s[sta[i]] == '('){fla = false;break;}else{break;}}if (fla){int x = 0;for (int i = top; i >= top; i--){if (s[sta[i]] == '#'){ans[sta[i]] += res;break;}}for (int i = 1; i <= top; i++){if (s[sta[i]] == '#')printf("%d\n", ans[sta[i]]);}}else{puts("-1");}
}

C o d e F o r c e s 977 F CodeForces 977F CodeForces977F C o n s e c u t i v e S u b s e q u e n c e Consecutive Subsequence ConsecutiveSubsequence
思路
用 m a p map map进行离散化 d p dp dp
参考代码

/** @Author: vain* @Date: 2020* @LastEditTime: 2020-10-12 21:26:12* @LastEditors: sueRimn* @Description: 学不会 dp 的 fw* @FilePath: \main\demo.cpp*/
#include <bits/stdc++.h>
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <ctime>
#include <cstring>
#include <cstdlib>
#include <math.h>
#include <bitset>
//#include <unordered_map>
using namespace std;
typedef long long ll;
#define ll long long
//typedef unsigned long long uint;
const int N = 5000 + 20;
const int maxn = 2e5 + 20;
const int mod = 998244353;
int cnt, head[maxn], pos[maxn];
//typedef pair<ll, ll> p;
//priority_queue<p, vector<p>, greater<p>> q;
//int sum[maxn];
double Max(double a, double b) { return a > b ? a : b; }
int min(int a, int b) { return a < b ? a : b; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a * b / gcd(a, b); }
void swap(int &x, int &y) { x ^= y, y ^= x, x ^= y; }
int lowbit(int x) { return (x) & (-x); }
//map<int, int> vis;
ll ksm(ll a, ll b)
{ll res = 1;while (b){if (b & 1)res = (res * a) % mod;b >>= 1, a = a * a % mod;}return res % mod;
}
int read(int &v)
{int k = 1;v = 0;int c = getchar();while (c < '0' || c > '9'){if (c == '-')k = 0;c = getchar();}while (c >= '0' && c <= '9')v = (v << 3) + (v << 1) + (c - 48), c = getchar();if (k == 0)v = -v;return c;
}
int sta[maxn], pre[maxn];
vector<int> f;
map<int, int> dp;
int main()
{int n;read(n);for (int i = 1; i <= n; i++)read(sta[i]), pre[i] = -1;for (int i = 1; i <= n; i++){dp[sta[i]] = max(dp[sta[i]],dp[sta[i] - 1] + 1);}int ans = 0, k;for (int i = 1; i <= n; i++){if (dp[sta[i]] > ans)ans = dp[sta[i]], k = i;}printf("%d\n", ans);int x = sta[k];for (int i = n; i >= 1; i--){if (sta[i] == x){f.push_back(i);x--;}}sort(f.begin(), f.end());for (int i = 0; i < f.size(); i++)printf("%d ", f[i]);puts("");
}

CodeForces 625A Guest From the Past

C o d e F o r c e s 279 C CodeForces 279C CodeForces279C L a d d e r Ladder Ladder
思路
差分构造+前缀和，当然有更简单的写法，我没有想到，太菜了

参考代码

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <ctime>
#include <cstring>
#include <cstdlib>
#include <math.h>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
using namespace std;
typedef long long ll;
#define ll long long
//pedef unsigned long long uint;
const int N = 5000 + 20;
const int maxn = 1e5 + 20;
const int mod = 998244353;
int cnt, head[maxn], pos[maxn];
//pedef pair<ll, ll> p;
//priority_queue<p, vector<p>, greater<p>> q;
ll ksm(ll a, ll b)
{ll res = 1;while (b){if (b & 1)res = (res * a) % mod;b >>= 1, a = a * a % mod;}return res % mod;
}
void read(int &v)
{int k = 1;v = 0;int c = getchar();while (c < '0' || c > '9'){if (c == '-')k = 0;c = getchar();}while (c >= '0' && c <= '9')v = (v << 3) + (v << 1) + (c - 48), c = getchar();if (k == 0)v = -v;
}
int d[maxn],dp[maxn],a[maxn],sam[maxn];
int main()
{int n,q;read(n),read(q);for(int i=1; i<=n; i++){read(a[i]);if(a[i]==a[i-1])sam[i]=sam[i-1]+1;if(a[i]>=a[i-1])d[i]=0;elsed[i]=1;}dp[0]=0;for(int i=1; i<=n; i++)dp[i]=dp[i-1]+d[i];while(q--){int l,r;read(l),read(r);if(dp[r]-dp[l]==0)puts("Yes");else if(dp[r]-dp[l]==r-l)puts("Yes");else{int x=dp[r]-dp[l];r=r-sam[r];if(dp[r]-dp[r-x]==x)puts("Yes");elseputs("No");}}return 0;
}

C o d e F o r c e s 476 C CodeForces 476C CodeForces476C D r e a m o o n a n d S u m s Dreamoon and Sums DreamoonandSums
思路
数学题，推导公式
参考代码

/** @Author: vain* @Date: 2020* @LastEditTime: 2020-10-11 16:49:16* @LastEditors: sueRimn* @Description: 学不会 dp 的 fw* @FilePath: \main\demo.cpp*/
#include <bits/stdc++.h>
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <ctime>
#include <cstring>
#include <cstdlib>
#include <math.h>
#include <bitset>
//#include <unordered_map>
using namespace std;
typedef long long ll;
#define ll long long
//typedef unsigned long long uint;
const int N = 5000 + 20;
const int maxn = 1e5 + 20;
const int mod = 1000000007;
//typedef pair<ll, ll> p;
//priority_queue<p, vector<p>, greater<p>> q;
//int sum[maxn];
double Max(double a, double b) { return a > b ? a : b; }
int min(int a, int b) { return a < b ? a : b; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a * b / gcd(a, b); }
void swap(int &x, int &y) { x ^= y, y ^= x, x ^= y; }
int lowbit(int x) { return (x) & (-x); }
//map<int, int> vis;
ll ksm(ll a, ll b)
{ll res = 1;while (b){if (b & 1)res = (res * a) % mod;b >>= 1, a = a * a % mod;}return res % mod;
}
int read(int &v)
{int k = 1;v = 0;int c = getchar();while (c < '0' || c > '9'){if (c == '-')k = 0;c = getchar();}while (c >= '0' && c <= '9')v = (v << 3) + (v << 1) + (c - 48), c = getchar();if (k == 0)v = -v;return c;
}
struct node
{int v, pos, nex, w;
} edge[maxn << 1];
int cnt, head[maxn], pre[maxn], vis[maxn];
int dwn[maxn], in[maxn], color[maxn];
vector<int> f;
void add(int u, int v, int w)
{edge[cnt].v = v, edge[cnt].w = w;edge[cnt].nex = head[u], head[u] = cnt++;
}
void dfs(int u, int fa)
{pre[u] = fa;for (int i = head[u]; ~i; i = edge[i].nex){int v = edge[i].v;if (v != fa){if (edge[i].w == 2)vis[v] = 1;dfs(v, u);}}
}
int main()
{ll a, b;scanf("%lld %lld", &a, &b);ll ans = 0, res = (b * (b - 1) / 2) % mod;for (int i = 1; i <= a; i++){ans = (ans + res * (i * b % mod + 1) % mod) % mod;}printf("%lld\n", ans);
}

CodeForces 833A The Meaningless Game
N e w Y e a r a n d t h e P e r m u t a t i o n C o n c a t e n a t i o n New Year and the Permutation Concatenation NewYearandthePermutationConcatenation
思路
打表找规律
参考代码

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <ctime>
#include <cstring>
#include <cstdlib>
#include <math.h>
#include <bitset>
//#include <unordered_map>
using namespace std;
typedef long long ll;
#define ll long long
//typedef unsigned long long uint;
const int N = 5000 + 20;
const int maxn = 1e5 + 20;
const int mod = 998244353;
//typedef pair<ll, ll> p;
//priority_queue<p, vector<p>, greater<p>> q;
//int sum[maxn];
ll ksm(ll a, ll b)
{ll res = 1;while (b){if (b & 1)res = (res * a) % mod;b >>= 1, a = a * a % mod;}return res % mod;
}
int read(int &v)
{int k = 1;v = 0;int c = getchar();while (c < '0' || c > '9'){if (c == '-')k = 0;c = getchar();}while (c >= '0' && c <= '9')v = (v << 3) + (v << 1) + (c - 48), c = getchar();if (k == 0)v = -v;return c;
}
int a[10];
vector<int>f;
ll dp[1000005];
int main()
{int n;ll sum=6;dp[1]=1,dp[2]=2,dp[3]=9;for(int i=4;i<=1000000;i++){sum=(sum*i)%mod;dp[i]=((((dp[i-1]-1)%mod)*i%mod)+sum)%mod;}read(n);printf("%lld\n",dp[n]);return 0;
}

C o d e F o r c e s 584 C CodeForces 584C CodeForces584C M a r i n a a n d V a s y a Marina and Vasya MarinaandVasya
思路
构造相同字符，模拟
参考代码

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <ctime>
#include <cstring>
#include <cstdlib>
#include <math.h>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
using namespace std;
typedef long long ll;
#define ll long long
const int N = 5000 + 20;
const int maxn = 1e5 + 20;
const int mod = 998244353;
//pedef pair<ll, ll> p;
//priority_queue<p, vector<p>, greater<p>> q;
ll ksm(ll a, ll b)
{ll res = 1;while (b){if (b & 1)res = (res * a) % mod;b >>= 1, a = a * a % mod;}return res % mod;
}
void read(int &v)
{int k = 1;v = 0;int c = getchar();while (c < '0' || c > '9'){if (c == '-')k = 0;c = getchar();}while (c >= '0' && c <= '9')v = (v << 3) + (v << 1) + (c - 48), c = getchar();if (k == 0)v = -v;
}
vector<int>samq,difq;
char s[maxn];
int check(char *s,string s1,string s2)
{int n=s1.size(),ans=0;for(int i=0; i<n; i++){if(s[i]!=s1[i]&&s[i]!=s2[i])ans++;}return ans;
}
int main()
{int n,t,ct;read(n),read(t);t=n-t;ct=t;string s1,s2;cin>>s1>>s2;for(int i=0; i<n; i++){if(s1[i]==s2[i])samq.push_back(i);elsedifq.push_back(i);}if(t==samq.size()){for(int i=0; i<difq.size(); i++){char c='a';int k=0;while(s1[difq[i]]==c||s2[difq[i]]==c){c=char(c+(++k)%26);}s[difq[i]]=c;}for(int i=0; i<samq.size(); i++){s[samq[i]]=s1[samq[i]];}printf("%s\n",s);}if(t<samq.size()){for(int i=0; i<difq.size(); i++){char c='a';int k=0;while(s1[difq[i]]==c||s2[difq[i]]==c){c=char(c+(++k)%26);}s[difq[i]]=c;}for(int i=0; i<samq.size(); i++){if(t){s[samq[i]]=s1[samq[i]];--t;}else{char c='a';int k=0;while(s1[samq[i]]==c){c=char(c+(++k)%26);}s[samq[i]]=c;}}printf("%s\n",s);}if(t>samq.size()){t-=samq.size();int t1=t,t2=t;int ok=0;for(int i=0; i<difq.size(); i++){char c='a';int k=0;if(t1||t2){if(ok%2==0&&t1){c=s1[difq[i]];t1--;}else if(t2){c=s2[difq[i]];t2--;}ok++;}else{while(s1[difq[i]]==c||s2[difq[i]]==c){c=char(c+((++k)%26));}}s[difq[i]]=c;}for(int i=0; i<samq.size(); i++){s[samq[i]]=s1[samq[i]];}if(t1||t2)puts("-1");elseprintf("%s\n",s);}return 0;
}

C o d e F o r c e s 598 D CodeForces 598D CodeForces598D I g o r I n t h e M u s e u m Igor In the Museum IgorIntheMuseum
思路
预处理连通块贡献
参考代码

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <ctime>
#include <cstring>
#include <cstdlib>
#include <math.h>
#include <bitset>
#include <unordered_map>
using namespace std;
typedef long long ll;
#define ll long long
const int N = 1000 + 20;
const int maxn = 1e5 + 5;
const int mod = 998244353;
//typedef pair<ll, ll> p;
//priority_queue<p, vector<p>, greater<p>> q;
ll ksm(ll a, ll b)
{ll res = 1;for (; b;){if (b & 1)res = (res * a) % mod;b >>= 1, a = a * a % mod;}return res;
}
int read(int &v)
{int k = 1;v = 0;int c = getchar();while (c < '0' || c > '9'){if (c == '-')k = 0;c = getchar();}while (c >= '0' && c <= '9')v = (v << 3) + (v << 1) + (c - 48), c = getchar();if (k == 0)v = -v;return c;
}
struct node
{int x,y;
} p;
vector<node>f[maxn*10];
int dir[4][2]= {1,0,0,1,0,-1,-1,0};
char mp[N][N];
int vis[N][N];
int ans[100005][2],corrt[1000005];
int ck[N][N];
int sum=0,n,m;
void check(int x,int y)
{if(x-1>0&&mp[x-1][y]=='*')sum++;if(y-1>0&&mp[x][y-1]=='*')sum++;if(x+1<=n&&mp[x+1][y]=='*')sum++;if(y+1<=m&&mp[x][y+1]=='*')sum++;
}
void dfs(int x,int y,int block)
{if(!vis[x][y]){node p;p.x=x,p.y=y;f[block].push_back(p);check(x,y);vis[x][y]=1;}else{return;}for(int i=0; i<4; i++){int xx=x+dir[i][0];int yy=y+dir[i][1];if(xx<0||yy<0||xx>n||yy>m||mp[xx][yy]=='*'||vis[xx][yy])continue;dfs(xx,yy,block);}
}
int main()
{int k;read(n),read(m),read(k);for(int i=1; i<=n; i++)scanf("%s",mp[i]+1);for(int i=1; i<=k; i++)scanf("%d %d",&ans[i][0],&ans[i][1]);int block=0;for(int i=1; i<=n; i++){for(int j=1; j<=m; j++){if((!vis[i][j])&&mp[i][j]=='.'){dfs(i,j,++block);corrt[block]=sum;sum=0;}}}for(int i=1;i<=block;i++){int l=f[i].size();for(int j=0;j<l;j++){ck[f[i][j].x][f[i][j].y]=corrt[i];}}for(int i=1;i<=k;i++){printf("%d\n",ck[ans[i][0]][ans[i][1]]);}return 0;
}

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