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农夫 John 余下了 m 批干草无法处理，他准备要开一个拍卖会去出售他的干草。现在有 n 个顾客，每个顾客的报价是 ai。现在 John
要确定一个单价，所有报价大于等于单价的顾客将会买到 1批干草（m 批甘草不用全卖完），总共获得的金钱作为收益。那么问题来了，如何设定单价，使得收益最大。
输入格式
第一行两个整数 m,n分别表示 m 批干草和 n 个顾客。
第二行 n个整数，ai 表示第 i 个顾客的报价。
数据范围：1≤n,m≤1000，1≤ai≤10000。
输出格式
两个用空格分隔的整数，分别表示单价和总收益。
如果有多个相等的最大收益，选取单价最小的那个。
AC代码如下：
#include<iostream> #include<cstdio> #include<algorithm> using namespace std;
int s[1009]; int main() { int n,m; scanf("%d %d",&m,&n); for(int i =
1;i<=n;i++) scanf("%d",&s[i]); sort(s+1,s+n+1,greater<int>()); int pr = 0; int
sum = 0; for(int i = 1;i<=n&&i<=m;i++) { if(sum<=s[i]*i) { pr = s[i]; sum =
s[i]*i; } } printf("%d %d\n",pr,sum); }
代码解析（题解）：
按照题目的意思进行枚举就可以
我们先排序（从大到小），这样做的好处：
以i顾客的报价ai为单价，收益为ai * i 。这样在枚举的过程中写的代码就更加舒服与简洁！！！
sort(s+1,s+n+1,greater<int>()); int pr = 0; int sum = 0; for(int i =
1;i<=n&&i<=m;i++) { if(sum<=s[i]*i) { pr = s[i]; sum = s[i]*i; } }
注意 ： if(sum<=s[i]*i)
因为题目的要求是：如果有多个相等的最大收益，选取单价最小的那个。
最后输出单价 与 收益 就ok了！
最后，感谢您的阅读！！！
备战蓝桥杯！！！
从今天开始，大一的我要再开一个专栏——备战蓝桥杯
希望大家可以多多支持！
同时也希望 备战蓝桥杯的同志们，金榜题名！！！
我不去想是否能够成功，既然选择了远方，便只顾风雨兼程！
