HDU1070 Milk 细节决定成败

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1070

注意：1.喝到第五天，第六天就不喝了  2.相同花费的，优先考虑容量大的  3.注意强制类型转换 4.精度一定要注意

附上题解：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 100 + 10;
typedef struct Milk{
    int yuan;
    int value;
    char name[maxn];
    double weight; //保存代价
}Milk;
Milk a[maxn]; //结构体数组存放

void solve(int n)
{
    int p, j;
    double min = 999999; //精度要注意
    for(int i = 0; i < n; i++){
        if(a[i].value < 200) continue;
        if(a[i].value >= 1000) p = 5;
        else p = a[i].value/200;
        a[i].weight = double(a[i].yuan/p); //注意类型转换
        if(a[i].weight < min){
            min = a[i].weight;
            j = i;
        }
        else if(a[i].weight == min){
            if(a[i].value > a[j].value){
                j = i;
            }
        }
    }
    printf("%s
", a[j].name);
}

int main()
{
    int t, n;
    while(~scanf("%d", &t)){
        while(t--){
            memset(a, 0, sizeof(a));
            scanf("%d", &n);
            for(int i = 0; i < n; i++)
                scanf("%s%d%d", a[i].name, &a[i].yuan, &a[i].value);
            solve(n);
        }
    }
    return 0;
}