ACM-ICPC 2018 沈阳赛区网络预赛 D F I G K持续更新中

/**
G. Spare Tire
链接:https://nanti.jisuanke.com/t/31448
题意:由递推式可得到数列的第n项为n*(n+1)，前n项和为n*(n+1)*(2*n+1)/6+n*(n+1)/2;
枚举m的质因子,通过容斥找出[1,n]中与m不互质的数
对于m的每个质因数组合y  可以得到最大和其不互质的数 m/y*y  最后依据容斥原理 奇加偶减冗余一下即可
*******tricks******
取余长度过长 wa....了一下午...O(1)直接ans wa
*/#include<bits/stdc++.h>
#define mod 1000000007
using namespace std;
typedef unsigned long long LL;const int maxn = 1e6 + 10;
LL arr[maxn];
int p;
LL inv6,inv2;
LL quick(LL x,LL y){LL ans= 1;while(y){if(y&1)ans=ans*x%mod;y=y>>1;x=x*x%mod;}return ans;
}
/**
4 4
9876543 1234567
*/LL get(LL x, LL y) {LL cnt = x / y;LL tmp=y*y%mod;tmp=tmp*cnt%mod*(cnt+1)%mod*(2*cnt+1)%mod*inv6;tmp=(tmp+y*cnt%mod*(cnt+1)%mod*inv2)%mod;return tmp;
}void getp(LL n) {  //分解质因子p = 0;for(int i = 2; i * i <= n; i++) {if(n % i == 0) {arr[p++] = i;while(n % i == 0) n /= i;}}if(n > 1) arr[p++] = n;
}int main() {LL n,m;inv6=quick(6ll,mod-2);inv2=quick(2ll,mod-2);while(~scanf("%lld %lld", &n,&m)) {getp(m);LL sum = n * (n + 1)%mod * (2 * n + 1)%mod*inv6%mod+n*(n+1)%mod*inv2%mod;LL ans = 0;for(int i = 1; i < (1 << p); i++) { //状压LL res = 0, cnt = 1;for(int j = 0; j < p; j++) {if(i & (1 << j)) {cnt *= arr[j];res++;}}if(res & 1) ans =(ans+get(n, cnt))%mod; //容斥else ans = (ans-get(n, cnt)+mod)%mod;}sum=(sum-ans+mod)%mod;printf("%lld\n",sum);}return 0;
}容斥的第二种写法;LL ans;
void dfs(int step,int flag,int tmp,ll x){if(step==p){ans=(ans+mod+flag*get(x,tmp))%mod;return ;}dfs(step+1,flag,tmp,x);dfs(step+1,-flag,lcm(tmp,arr[step]),x);
}int main() {LL n,m;inv6=quick(6ll,mod-2);inv2=quick(2ll,mod-2);while(~scanf("%lld %lld", &n,&m)) {getp(m);ans = 0;dfs(0,1,1,n);printf("%lld\n",ans);}return 0;
}

/**F ：Fantastic Graph
链接:https://nanti.jisuanke.com/t/31447
最大流模板题;
*/#include<cstdio>
#include<cstring>
using namespace std;
const int inf=1<<30;
const int nmax=100105;//点的数量
const int mmax=3000005;//边的数量
struct node
{int c,u,v,next;void insert(int nu,int nv,int nc,int nnext){u=nu;v=nv;c=nc;next=nnext;}
}edge[mmax];
int cnt,head[nmax];
int cur[nmax],ps[nmax],dep[nmax];void addedge(int u,int v,int w) //加边
{edge[cnt].insert(u,v,w,head[u]);head[u]=cnt++;edge[cnt].insert(v,u,0,head[v]);head[v]=cnt++;
}int dinic(int s, int t){                       //  dinicint tr, res = 0;int i, j, k, f, r, top;while(1){memset(dep, -1, sizeof(dep));for(f = dep[ps[0]=s] = 0, r = 1; f != r;){for(i = ps[f ++], j = head[i]; j; j = edge[j].next){if(edge[j].c && dep[k=edge[j].v] == -1){dep[k] = dep[i] + 1;ps[r ++] = k;if(k == t){f = r; break;}}}}if(dep[t] == -1) break;memcpy(cur, head, sizeof(cur));i = s, top = 0;while(1){if(i == t){for(tr =inf, k = 0; k < top; k ++){if(edge[ps[k]].c < tr){tr = edge[ps[f=k]].c;}}for(k = 0; k < top; k ++){edge[ps[k]].c -= tr;edge[ps[k]^1].c += tr;}i = edge[ps[top=f]].u;res += tr;}for(j = cur[i]; cur[i]; j = cur[i] =edge[cur[i]].next){if(edge[j].c && dep[i]+1 == dep[edge[j].v]){break;}}if(cur[i]){ps[top ++] = cur[i];i = edge[cur[i]].v;}else{if(top == 0){break;}dep[i] = -1;i = edge[ps[-- top]].u;}}}return res;
}int main()
{int l,r,m,x,y,s,t,S,T,u,v,ooo=1;while(~scanf("%d%d%d",&l,&r,&m)){printf("Case %d: ",ooo++);cnt=2;memset(head,0,sizeof(head));scanf("%d%d",&x,&y);while(m--){scanf("%d%d",&u,&v);addedge(u,v+l,1);}s=0,t=l+r+1;S=l+r+2,T=l+r+3;for(int i=1;i<=l;i++){addedge(s,i,y-x);addedge(S,i,x);addedge(s,T,x);}addedge(t,s,inf);for(int i=1;i<=r;i++){addedge(i+l,t,y-x);addedge(S,t,x);addedge(i+l,T,x);}if(dinic(S,T)==(l+r)*x)printf("Yes\n");else printf("No\n");}
}

/**
I Lattice's basics in digital electronics
链接:https://nanti.jisuanke.com/t/31450
题意:可仔细阅读case2 的模拟过程;
大模拟.....先进行转换后转换回去,处理好合法串的情况,注意查询输出即可;
********tricks******
str+=s the time << str=str+s;
赛中并没有看到这个题,遗憾;
*/#include<bits/stdc++.h>
#define ll long long
using namespace std;map<string,int>mp1;
map<char,string>mp2;
string s1,str,s2,s3,tmp;int main (){mp2['0']="0000";mp2['1']="0001";mp2['2']="0010";mp2['3']="0011";mp2['4']="0100";mp2['5']="0101";mp2['6']="0110";mp2['7']="0111";mp2['8']="1000";mp2['9']="1001";mp2['A']="1010";mp2['B']="1011";mp2['C']="1100";mp2['D']="1101";mp2['E']="1110";mp2['F']="1111";mp2['a']="1010";mp2['b']="1011";mp2['c']="1100";mp2['d']="1101";mp2['e']="1110";mp2['f']="1111";int t;scanf("%d",&t);int x;while(t--){int n,m,cnt2;scanf("%d %d",&n,&m);mp1.clear();s1="";s2="";tmp="";for(int i=0;i<m;i++){cin>>x>>str;mp1[str]=x;}cnt2=0;cin>>str;int l=str.length();for(int i=0;i<l;i++) s1+=mp2[str[i]];for(int i=0;i<4*l;i+=9){s3=s1.substr(i,9);if(s3.length()!=9) break;int cnt=0;for(int j=0;j<8;j++) {if(s3[j]=='1') cnt++;}if((cnt%2==1)&&s1[i+8]=='0') s2+=s3.substr(0,8);if((cnt%2==0)&&s1[i+8]=='1') s2+=s3.substr(0,8);s3="";}int len=s2.length();int k=0;for(int i=0;i<len;i++){tmp+=s2[i];if(mp1[tmp]) {int x=mp1[tmp];printf("%c",(char)x);tmp="";k++;}if(k==n) break;}cout<<endl;}return 0;
}

/***K. Supreme Number
链接:https://nanti.jisuanke.com/t/31452
题意:求1-->n最大的数&&该数的子序列组合后为质数;
打表即可;;
******tricks****
基本语法;......
*/#include<bits/stdc++.h>
#define ll long long
using namespace std;int a[100]={2,3,5,7,11,13,17,23,31,37,53,71,73,113,131,137,173,311,317};int main (){ios_base::sync_with_stdio(false);cin.tie(0);int t;cin>>t;string str;for(int cas=1;cas<=t;cas++){cin>>str;int len=str.length();if(len>=4) printf("Case #%d: 317\n",cas);else {int ans=0,len=str.length();for(int i=0;i<len;i++) ans=ans*10+(str[i]-'0');if(ans>=317) { printf("Case #%d: 317\n",cas);}else for(int i=0;i<=19;i++){if(a[i]>ans) { printf("Case #%d: %d\n",cas,a[i-1]);break;}else if(a[i]==ans) { printf("Case #%d: %d\n",cas,a[i]);break;}}}}return 0;
}

/**
D. Made In Heaven
链接:https://nanti.jisuanke.com/t/31445
题意:是否存在第k短路且长度是否小于等于T.
思路:*A + dijstra +堆优化 模板题,原题;
*/#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;const int maxn = 10000 + 5;
const int INF = 0x3f3f3f3f;
int s, t, k;bool vis[maxn];
int dist[maxn];struct Node {int v, c;Node (int _v = 0, int _c = 0) : v(_v), c(_c) {}bool operator < (const Node &rhs) const {return c + dist[v] > rhs.c + dist[rhs.v];}
};struct Edge {int v, cost;Edge (int _v = 0, int _cost = 0) : v(_v), cost(_cost) {}
};vector<Edge>E[maxn], revE[maxn];void Dijkstra(int n, int s) {memset(vis, false, sizeof(vis));for (int i = 1; i <= n; i++) dist[i] = INF;priority_queue<Node>que;dist[s] = 0;que.push(Node(s, 0));while (!que.empty()) {Node tep = que.top(); que.pop();int u = tep.v;if (vis[u]) continue;vis[u] = true;for (int i = 0; i < (int)E[u].size(); i++) {int v = E[u][i].v;int cost = E[u][i].cost;if (!vis[v] && dist[v] > dist[u] + cost) {dist[v] = dist[u] + cost;que.push(Node(v, dist[v]));}}}
}int astar(int s) {priority_queue<Node> que;que.push(Node(s, 0)); k--;while (!que.empty()) {Node pre = que.top(); que.pop();int u = pre.v;if (u == t) {if (k) k--;else return pre.c;}for (int i = 0; i < (int)revE[u].size(); i++) {int v = revE[u][i].v;int c = revE[u][i].cost;que.push(Node(v, pre.c + c));}}return -1;
}void addedge(int u, int v, int w) {revE[u].push_back(Edge(v, w));E[v].push_back(Edge(u, w));
}int main() {int n, m, u, v, w,yyy;while (scanf("%d%d", &n, &m) != EOF) {for (int i = 0; i <= n; i++) {E[i].clear();revE[i].clear();}scanf("%d%d%d%d", &s, &t, &k,&yyy);for (int i = 0; i < m; i++) {scanf("%d%d%d", &u, &v, &w);addedge(u, v, w);}Dijkstra(n, t);if (dist[s] == INF) {puts("Whitesnake!");continue;}if (s == t) k++;if(astar(s)<=yyy)puts("yareyaredawa");else puts("Whitesnake!");}return 0;
}

ACM-ICPC 2018 沈阳赛区网络预赛 D F I G K持续更新中相关推荐

ACM-ICPC 2018 沈阳赛区网络预赛 Spare Tire（容斥+公式推）
A sequence of integer \lbrace a_n \rbrace{an} can be expressed as: \displaystyle a_n = \left\{ \beg ...
ACM-ICPC 2018 沈阳赛区网络预赛（E F G J K)
ACM-ICPC 2018 沈阳赛区网络预赛(E F G J K) 复杂的模拟题懒癌患者表示写不动 D. Made In Heaven (K短路) 略 int head[MAXN]; int cure ...
ACM-ICPC 2018 沈阳赛区网络预赛 F. Fantastic Graph(有源上下界最大流模板)
关于有源上下界最大流: https://blog.csdn.net/regina8023/article/details/45815023 #include<cstdio> #includ ...
ACM-ICPC 2018 沈阳赛区网络预赛 D Made In Heaven（第k短路，A*算法）
https://nanti.jisuanke.com/t/31445 题意能否在t时间内把第k短路走完. 分析 A*算法板子. #include <iostream> #include ...
【ACM-ICPC 2018 沈阳赛区网络预赛 I】Lattice's basics in digital electronics
[链接] 我是链接,点我呀:) [题意] [题解] 每个单词的前缀都不同. 不能更明示了... 裸的字典树. 模拟一下.输出一下就ojbk了. [代码] #include <bits/stdc+ ...
Convex Hull (ACM-ICPC 2018 沈阳赛区网络预赛) 存个公式
Convex Hull gay(i)={0ifi=k×x×x,x>k,k>1i×ielse}求∑i=1n∑j=1igay(j)=∑i=1n(n−i+1)gay(i)=∑i=1n(n−i+1 ...
ACM-ICPC 2018 沈阳赛区网络预赛 J Ka Chang 分块
https://nanti.jisuanke.com/t/31451 对每层的个数分块当这个深度的节点个数>block时暴力维护每个点的子树有多少个这个深度的节点这样的层数最多有n/blo ...
ACM-ICPC 2018 沈阳赛区网络预赛 J Ka Chang（树分块）
思路因为不同深度的节点数量不同,数量少的节点,可以考虑直接进行单点更新,对于数量多的节点,可以直接记录这一层增加的值,查询的时候,看每一层有多少个节点,最后乘上增加的值就行了. 具体实现先设定一个 ...
【ACM-ICPC 2018 沈阳赛区网络预赛】I.Lattice's basics in digital electronics ---- 字典树
题目传送门做法: 用字典树存好译码词,然后模拟即可 AC代码: #include <bits/stdc++.h> using namespace std;#define IO ios_b ...
ACM-ICPC 2018 沈阳赛区网络预赛 G. Spare Tire
原题传送门题意: 给定一个函数a(n) ,和两个整数,n, m; 求闭区间[1, n]内,所有的a(x)的和, x满足gcd(x, m) == 1 思路: (下午打比赛的时候思路还是蛮接近的,但就是 ...

ACM-ICPC 2018 沈阳赛区网络预赛 D F I G K持续更新中

ACM-ICPC 2018 沈阳赛区网络预赛 D F I G K持续更新中相关推荐

最新文章

热门文章