Educational Codeforces Round 94 (Rated for Div. 2)

这次做了ABCD，今天下午就要上学去了溜了溜了，早上起来补的题解。

A - String Similarity

分析可知可构造w[i]=s[2*i]即可满足题意

#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{IO;int T;cin>>T;while(T--){int n;cin>>n;string s;cin>>s;string w;for(int i=0;i<n;i++) w+=s[2*i];cout<<w<<endl;}return 0;
}

B - RPG Protagonist

枚举+贪心
这题很有意思，昨天做的时候群友许多都说C比B简单，不过我B是1A，而Cwa2次
刚开始想了很久确实没啥思路，但是一看数据范围2⋅1052·10^52⋅105好像直接可以枚举数量。对于题目不妨设s≥ws \ge ws≥w，如果s<ws<ws<w可以交换一下，从集合角度考虑，最优解一定是一个人拿了aaa个sss，bbb个www，另一个人拿了ccc个sss，ddd个www，因此可以考虑枚举第一个人拿sss的数目，然后b,c,db,c,db,c,d贪心可以算出来。

#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
ll p,f;
ll cntw,cnts;
ll s,w;
int main()
{IO;int T;cin>>T;while(T--){cin>>p>>f;cin>>cnts>>cntw;cin>>s>>w;if(s<w) swap(s,w),swap(cnts,cntw);ll maxs=p/s;ll res=0;for(int i=0;i<=min(maxs,cnts);i++){ll a=i,b=min(cntw,(p-i*s)/w);ll d=min(f/w,cntw-b);ll c=min((f-d*w)/s,cnts-i);res=max(res,a+b+c+d);}cout<<res<<endl;}return 0;
}

C - Binary String Reconstruction

这题看错题了我晕

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Educational Codeforces Round 94 (Rated for Div. 2)

A - String Similarity

B - RPG Protagonist

C - Binary String Reconstruction

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