E. Expenditure Reduction

从左右往右找到包含B字符的最近位置,然后从这个位置有从右到左找回去找到包含完所有B字符的位置,这个区间就是答案

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define endl '\n'
#define int long long

using namespace std;

const int N = 1e6+10, M = 998244353;

//typedef long long ll;
typedef pair<int,int> PII;
int n,m,t,k;
map<int,int> mp;
priority_queue<int> QQ;
deque<int> Q;
int ksm(int a,int b,int mod){
    int res = 1;
    a %= mod;
    while(b){
        if(b & 1)
            res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}
void solve() {
   string a,b;
   cin >> a >> b;
   int l = 0, r = a.size() - 1, pos = 0;
   for(int i = 0;i < a.size();i++){
       if(a[i] == b[pos])
           pos++;
       if(pos == b.size()){
           r = i;
           break;
       }
   }
   pos --;
   //cout << pos << endl;
   for(int i = r;i >= 0;i--){
       if(a[i] == b[pos])
           pos--;
       if(pos == -1){
           l = i;
           break;
       }
   }
  //cout << l << ' ' << r << endl;
   cout << a.substr(l, r - l + 1) << endl;
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int Ke_scholar = 1;
    cin >> Ke_scholar;
    while(Ke_scholar--)
        solve();
    return 0;
}
/*

 */

G. Gua!

签到题,注意细节.

#include<bits/stdc++.h>

//#define int long long
#define endl '\n'
#define PII pair<int,int>

using namespace std;

const int N = 2010,mod = 1e9 + 7;
int  n,s,v,m;

void solve() {
    int b, r, d, s;
    cin >> b >> r >> d >> s;
    if (b == 0||r==0) {
        if (d)cout << "gua!" << endl;
        else cout << "ok" << endl;
        return;
    }
    int x = (d + b - 1) / b;
    int y = r * s / 60 + 1;
    //cout << x << ' ' << y << endl;
    if (x <= y)cout << "ok" << endl;
    else cout << "gua!" << endl;
}

int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    int Ke_scholar = 1;
    cin >> Ke_scholar;
    while(Ke_scholar--)solve();
    return 0;
}

H. Heirloom Painting

考虑最后一次涂色，必然会形成一个长度不小于