CCCC 月饼

https://www.patest.cn/contests/gplt/L2-003

题解：按平均值贪心。

坑：有一个样例卡住了，是因为

while (i<=n&&x - bs[i].num >= 0) 
{i++}
形如这样的循环结构要防止bs越界（i++以后再次访问了bs[i])

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <math.h>
#include <string.h>
#include <string>
#include <map>
#include<stack>
#include<set>
#include<string.h>
#define pb push_back
#define _for(i, a, b) for (int i = (a); i<(b); ++i)
#define _rep(i, a, b) for (int i = (a); i <= (b); ++i)

using namespace std;
const  int N = 10000 + 5;
//double num[N], price[N], ave[N];
struct bing {
    float num, price, ave;
}bs[N];
bool cmp(bing a, bing b) {
    return a.ave > b.ave;
}
int main() {
    int n;
    cin >> n;
    float x; cin >> x;
    for (int i = 1; i <= n; i++) {
        cin >> bs[i].num;
    }
    for (int i = 1; i <= n; i++) {
        cin >> bs[i].price;
        bs[i].ave = bs[i].price / bs[i].num;
        

    }
    sort(bs + 1, bs + 1 +    n, cmp);
    float ans = 0;
    int i = 1;
    while (i<=n&&x - bs[i].num >= 0) {
        x -= bs[i].num;
        ans += bs[i].price;
        i++;
    }
    if(i<=n)ans += bs[i].ave*x;
    printf("%.2lf", ans);
    system("pause");


}