字节跳动劝退题

老哥第三题的输入该怎么处理呢？求问第三题

求大佬解答最后一题

第二题是先把当前题目和之前的题目的时间求和，如果小于等于规定时间，直接返回0，如果大于的话，把之前的题目时间从小到大排序，从最大的时间开始减，直到时间总和小于等于规定时间，然后返回减掉的个数

大佬，第二题求个思路。我用大根堆做，在Eclipse上感觉好好的，一提交 0%……

贴下第二题代码： #include<iostream> #include<vector> #include<queue> #include<iostream> using namespace std; int Get_Num(priority_queue<int> Prac,int sum,int total) {     int ans = 0;     while(total < sum)     {         sum -= Prac.top();         Prac.pop();         ans++;     }     return ans; } int main() {     int T;     cin>>T;     while(T--)     {         int n,m;         cin>>n>>m;         int sum = 0;         priority_queue<int> Prac;         for(auto i = 0;i < n;i++)         {             int pi;             cin>>pi;             if(sum+pi <= m)             {                 cout<<0<<" ";                 sum += pi;                 Prac.push(pi);                              }else             {                 int total = m - pi;                 cout<<Get_Num(Prac,sum,total)<<' ';                 sum+=pi;                 Prac.push(pi);             }         }         cout<<endl;     }     return 0; } 有人能帮我看一下我的第一题嘛： #include<string> #include<vector> #include<iostream> using namespace std; int main() {     int n;     cin>>n;     string S;     cin>>S;     vector<int> lr(n,0);     vector<int> cnt(n,0);     for(auto i = 0 ; i < n; i ++)     {         if(i == 0)         {             if(S[i] == 'O')             {                 cnt[i] = 0;             }else             {                 int j = i+1;                 while(S[j] != 'O') j++;                 lr[i] = 1;                 cnt[i] = j-i;             }         }         else if(i == n-1)         {             if(S[i] == 'O')             {                 cnt[i] = 0;             }             else              {                 cnt[i] = cnt[i-1]+1;             }         }         else         {             if(S[i] == 'O')             {                 lr[i]=0;                 cnt[i] = 0;             }else if(lr[i-1] == 1)             {                 lr[i] = 1;                 cnt[i] = cnt[i-1] -1;             }else if(lr[i-1] == 0)             {                 if(S[i+1] == 'O')                 {                     lr[i] = 1;                 }                 else                 {                     lr[i] = -1;                 }                 cnt[i] = 1;             }             else             {                 bool right = false;                 for(auto j = cnt[i-1]-1;j <= cnt[i-1]+1;j++)                 {                     if(S[i+j] == 'O')                     {                         right = true;                         cnt[i] = j;                     }                 }                 if(right == true)                 {                     lr[i] = 1;                 }                 else                  {                     lr[i] = -1;                     cnt[i] = cnt[i-1]+1;                 }             }         }     }     for(auto I:cnt)     {         cout<<I<<' ';     }          return 0; }  第二题全过了，第一题只过了10%

100, 100, 90, 8 太菜了

第二道题思想是总时间减当前时间得剩余时间，再按照前面从小到大排序求和，直到求的和大于剩余时间，统计已求和的个数，再前面总个数减已求和个数就是答案，是这个思路吗？

0.95，0.28，0.7，0

可以求个第二题吗

%6是咋做到的😂

第二道不知道为什么过不去