CF EDU94 div2

A 

t组测试，字符串s的长度 2 * n - 1

输出长度为n的字符串w,满足s[1,n] s[2,n + 1],s[3,n + 2],s[n,2 * n - 1]相似

从第一个字母开始去，隔着一个再取下一个

思维

#include <bits/stdc++.h>
using namespace std;
int t,n;
string s;
int main() {
    //freopen("in","r",stdin);
    ios::sync_with_stdio(0);
    cin >> t;
    while(t--){
        cin >> n >> s;
        for(int i = 0; i < s.size(); i+=2)
            cout << s[i];
        cout << endl;
    }
    return 0;
}

B

t组测试样例。你有p，你的追随者有f,sword和axe数量分别是cnts,cntw;

每个分别消耗s,w;求获得的weapon数量最大值

设

	p	f
s	s1	s2
w	w1	w2

由题意满足条件

s1 + s2 <= cnts, w1 + w2<= cntw;
s1 * s + w1 * w<= p; s2 * s + w2 * w<= f;

先假设 s  < w; s1开始枚举

枚举

#include <bits/stdc++.h>
using namespace std;
int t,p,f,cnts,cntw,s,w;
int ans,s1,w1,s2,w2;
int main() {
   // freopen("in","r",stdin);
    ios::sync_with_stdio(0);
    cin >> t;
    while(t--){
        cin >> p >> f >> cnts >> cntw >> s >> w;
        ans = 0;
        if(s > w){swap(s,w);swap(cnts,cntw);}
        for(s1 = 0; s1 <= cnts && s1 * s <= p; s1 ++){
            w1 = min(cntw,(p - s1 * s) / w );
            s2 = min(cnts - s1,f / s);
            w2 = min(cntw - w1,(f - s * s2 ) / w);
            ans = max(ans,s1 + w1 + s2 + w2);
        }
        cout << ans << endl;
    }
    return 0;
}

C

大佬的博客

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int t,x,flag,len;
string S;
int w[maxn],s[maxn];
int main() {
   // freopen("in","r",stdin);
    ios::sync_with_stdio(0);
    cin >> t;
    while(t--) {
        cin >> S >> x;
        int i;
        len = S.size();
        for (i = 0; i < len; i++) {
            s[i + 1] = S[i] - '0';w[i + 1] = 1;
        }
        for (i = 1; i <= len; i++) {
            if(!s[i]){
                if(i - x > 0) w[i - x] = 0;
                if(i + x <= len) w[i + x] = 0;
            }
        }
        flag = 1;
        for (i = 1; i <= len; i++) {
          if(s[i]){
              if(((i - x > 0 && !w[i - x]) || i - x <= 0) && ((i + x <= len && !w[i + x])|| i + x > len)){
                  flag = 0;
                  break;
              }
          }
        }
        if (flag) for (i = 1; i <= len; i++) cout << w[i];
        else cout << -1;
        cout << endl;
    }
    return 0;
}