Problem Statement

There are N dishes, and the ii-th dish has a sweetness of Ai​ and a saltiness of Bi​.

Takahashi plans to arrange these N dishes in any order he likes and eat them in that order.

He will eat the dishes in the arranged order, but he will stop eating as soon as the total sweetness of the dishes he has eaten exceeds X or the total saltiness exceeds Y.

Find the minimum possible number of dishes that he will end up eating.

 

Constraints

• 1≤N≤2×105
• 1≤X,Y≤2×1014
• 1≤Ai,Bi≤109
• All input values are integers.

原题链接：
 

https://atcoder.jp/contests/abc364/tasks/abc364_c

思路讲解：
 

题干要求使其吃尽量少的菜，他一但吃的甜大于X 或者 吃的咸大于Y ，那么他将不能吃了，

因此我们让其按咸从大到小排序，让其按甜从大到小排序，看看哪种情况下他吃的菜的数目最少即可。

代码实现：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>using namespace std;int main()
{int N; cin>>N;vector<int> A(N);vector<int> B(N);int ans=N;long long X,Y; cin>>X>>Y;for(int i=0;i<N;i++) cin>>A[i];for(int i=0;i<N;i++) cin>>B[i];sort(A.begin(),A.end(),greater<>());sort(B.begin(),B.end(),greater<>());for(int i=0;i<N;i++){X-=A[i];if(X<0){ans=min(ans,i+1);}}for(int i=0;i<N;i++){Y-=B[i];if(Y<0){ans=min(ans,i+1);}}cout<<ans<<endl;return 0;
}