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\n", 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;

}