【Codeforces Round #544 (Div. 3)】 A B C D E F1 F2

这场A是最失误的地方能int读入却用char数组读入感觉失策了

题意给你开始时间和结尾时间问你这两个时间的中间时间是什么

做法做法显然骚操作在此

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endlint main()
{int t1,f1,t2,f2;scanf("%d:%d",&t1,&f1);scanf("%d:%d",&t2,&f2);int tim1 = t1*60+ f1;int tim2 = t2*60 + f2;int jian = (tim2 - tim1)/2;tim1 += jian;printf("%02d:%02d\n",tim1/60,tim1%60);return 0;
}

题意给你n个数问你能分成几组能整除k

做法我们把所有数mod k 那么就得到k 组数就可以通过性质做了注意 i 如果等于 k-i 则需要取一般而不是min

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
int a[200025],cnt[200];int main()
{int n,k;scanf("%d%d",&n,&k);for(int i = 1;i<=n;++i)scanf("%d",&a[i]),a[i]%=k,cnt[a[i]]++;long long ans = 0;for(int i = 0;i<k/2+1;i++){if(i==0){ans+=cnt[i]/2;cnt[i]%=2;}else{if(k-i==i){ans+=cnt[i]/2;cnt[i]%=2;}else{long long minn = min(cnt[i],cnt[k-i]);ans = ans+minn;//dbg(minn);cnt[i]-=minn;cnt[k-i]-=minn;}}}printf("%lld\n",ans*2);return 0;
}

题意给你n个人的能力值问你分组使得最大能力值的人减去最小能力值的人差不大于5 一组最多有多少人

做法尺取即可

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
const int MAX_N = 200025;
int arr[MAX_N];
int main()
{int n,minn,maxx,ans=1;scanf("%d",&n);for(int i = 1;i<=n;++i)  scanf("%d",&arr[i]);sort(arr+1,arr+1+n);for (int l=1,r=1,now=0;l<=n;){while(arr[r]-arr[l]<=5&&r<=n){r++;}ans = max(r-l,ans);l++;}printf("%d\n",ans);return 0;
}

题意给你n组 a[i] 和 b[i] 问你使得 k*a[i] + b[i] = 0 同时满足条件最多的时候k是多少

做法一开始的做法是b[i]/a[i] 存在map里面找最大注意如果a[i] 等于 0 此时分两种如果a[i] b[i] 同时为0 那么另外记录tmp++(最终答案加上tmp) 如果a[i] 为0 b[i]不为0 则无法实现一开始会丢精度尤其是

999999999 999999998

1000000000 999999999

这组数据后来改成long double 过了实际上想到了gcd分解但是没有处理好负号这题用一个 pair的map存是一个小trick

值得大家学习

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
const int MAX_N = 200025;
double arr[MAX_N],arr_[MAX_N];
map<long double,int> mp;
int main()
{int n,ans = 0,cnt = 0;scanf("%d",&n);for(int i =1;i<=n;++i){scanf("%lf",&arr[i]);}for(int i = 1;i<=n;++i){scanf("%lf",&arr_[i]);if(arr[i]==0&&arr_[i]!=0) continue;if(arr[i]==0&&arr_[i]==0){cnt++;continue;}//printf("%.15Lf\n",(long double)arr_[i]/arr[i]);mp[(long double)(long double)arr_[i]/arr[i]]++;ans = max(ans,mp[(long double)arr_[i]/arr[i]]);}printf("%d\n",ans+cnt);return 0;
}

延续着C题的题意问你能组成那么多符合的组取k个会有多少人被组

首先排个序这样就可以二分了

dp[i][j] 代表前i个人已经组了k组的情况容易想到转移式为

dp[i][j] =max(dp[i-1][j],dp[i][j-1],dp[x-1][j-1]+i-x+1) x为第i个作为最高值组队的最小值的坐标

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
const int MAX_N = 5040;
int a[MAX_N],dp[MAX_N][MAX_N];
int main()
{int n,k;scanf("%d%d",&n,&k);for(int i = 1;i<=n;++i){scanf("%d",&a[i]);}sort(a+1,a+1+n);for(int i = 1;i<=n;++i){for(int j = 1;j<=min(i,k);j++){dp[i][j] = max(dp[i][j-1],dp[i-1][j]);int pos = lower_bound(a+1,a+1+n,a[i]-5)-a;dp[i][j] = max(dp[i][j],dp[pos-1][j-1]+i-pos+1);}}printf("%d\n",dp[n][k]);return 0;
}

题意给你n个点 m条边问你能输出一种生成树(保证存在) 使得最大度数点最大

做法我们把最大度数点找到然后连边把剩下的点跑一个最小生成树即可

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
const int MAX_N = 200025;
const int MAX_M = 400025;
int eid ,fa[MAX_N],a[MAX_N],cnt[MAX_N],b[MAX_N];
vector<int > vt[MAX_N];
struct node_
{int x,y;
}ans_[MAX_N];
struct node
{int u,v,w;bool operator< (const node &a) const{return w < a.w;}
}e[MAX_M];int get(int x)
{if(fa[x]==x) return x;return fa[x] = get(fa[x]);
}
void Merge(int x,int y)
{x = get(x),y = get(y);if(x!=y){fa[x] = y;}return ;
}
int main()
{int n,m,maxx =-1,maxx_i;scanf("%d%d",&n,&m);for(int i = 1;i<=m;++i){scanf("%d%d",&a[i],&b[i]);e[eid].u = a[i];e[eid++].v = b[i];e[eid].u = b[i];e[eid++].v = a[i];vt[a[i]].push_back(b[i]);vt[b[i]].push_back(a[i]);cnt[a[i]]++;cnt[b[i]]++;}for(int i = 1;i<=n;++i){if(cnt[i]>maxx){maxx = cnt[i];maxx_i = i;}}int sz = vt[maxx_i].size();for(int i = 1;i<=n;++i)fa[i] = i;int mst = n;for(int i = 0;i<sz;++i){int x = get(maxx_i),y = get(vt[maxx_i][i]);ans_[n-mst].x = maxx_i;ans_[n-mst].y = vt[maxx_i][i];mst--;Merge(x,y);}for(int i = 0;i<eid&&mst>1;i++){int x = get(e[i].u),y = get(e[i].v);if(x==y) continue;if(x!=y){ans_[n-mst].x = e[i].u;ans_[n-mst].y = e[i].v;Merge(x,y);mst--;}}for(int i = 0;i<n-1;i++)printf("%d %d\n",ans_[i].x,ans_[i].y);return 0;
}

接着上一题不过有个k 就是问你1这个点需要有k的度数

我们怎么做呢？首先排除a 处理出所有的联通块如果联通块个数大于k 那么就是NO 如果小于k 我们要把所有联通块和1连边然后再连和1相连的边知道k为0 然后再把所有联通快连边即可

代码实现的顺序分别是先连1和联通快 -> 再补度数 ->再连联通快内部的

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
const int MAX_N = 200025;
int tot;
vector<int > one[MAX_N],dsu[MAX_N];
bool vis[MAX_N];
vector<pair<int ,int > > ans;
vector<int > vt[MAX_N];
void dfs(int x)
{vis[x] = true;dsu[tot].push_back(x);int sz = vt[x].size();for(int i = 0;i<sz;++i){if(vis[vt[x][i]]) continue;if(vt[x][i]==1){one[tot].push_back(x);continue;}dfs(vt[x][i]);}
}
int main()
{int n,m,k,a,b,cnt = 0;scanf("%d%d%d",&n,&m,&k);for(int i = 1;i<=m;++i){scanf("%d%d",&a,&b);if(a==1||b==1) cnt++;vt[a].push_back(b);vt[b].push_back(a);}if(cnt<k){printf("NO\n");return 0;}for(int i = 2;i<=n;++i){if(!vis[i]){tot++;dfs(i);}}if(tot>k){printf("NO\n");return 0;}queue<int > q;memset(vis,0,sizeof(vis));vis[1] = true;for(int i = 1;i<=tot;i++){//dbg(one[i].back());ans.push_back(make_pair(1,one[i].back()));vis[one[i].back()] = true;q.push(one[i].back());one[i].pop_back();k--;}for(int i = 1;i<=tot;++i){while(k>0&&one[i].size()){ans.push_back(make_pair(1,one[i].back()));vis[one[i].back()] = true;q.push(one[i].back());one[i].pop_back();k--;}}while(!q.empty()){int tmp = q.front();//dbg(tmp);q.pop();for(int i = 0;i<vt[tmp].size();i++){if(!vis[vt[tmp][i]]){//dbg2(tmp,vt[tmp][i]);vis[vt[tmp][i]] = true;ans.push_back(make_pair(tmp,vt[tmp][i]));q.push(vt[tmp][i]);}}}printf("YES\n");for(int i = 0;i<n-1;i++)printf("%d %d\n",ans[i].first,ans[i].second);return 0;
}

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