HDOJ 1070 Milk

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【题意】喝牛奶，

1. 每天喝200ml。

2. 最多喝5天，超过5天剩下多少也不喝了。

3. 剩下的如果小于一天喝的（200ml）就不要了。

4. 所有的牛奶都是当天产的。

5. 按照每天的价格选择最便宜的。

6. 如果价格都是最便宜的，选择大瓶（容量大）的。

【代码：WA】：考虑错了2,5,6条件。

#include <iostream>#include <iomanip>#include <cstring>#include <cstdlib>#include <cstdio>using namespace std;#define MAX 100+10struct Milk{    char name[MAX];    int price;    int volumn;};int cmp(const void *a, const void *b){    struct Milk *c = (Milk*)a;    struct Milk *d = (Milk*)b;    if ((c->price/c->volumn) == (d->price/d->volumn))        return strlen(c->name) - strlen(d->name);    else        return (c->volumn/c->price) - (d->volumn/d->price);}int main(){    int N = 0;    cin >> N;    while (N--)    {        Milk milk[MAX];        int n = 0, i = 0;        cin >> n;        for (i = 0; i < n; )        {            Milk temp;            cin >> temp.name >> temp.price >> temp.volumn;            if (temp.volumn >= 200)                milk[i++] = temp;            else                n--;        }        qsort(milk, n, sizeof(milk[i]), cmp);        cout << milk[n-1].name << endl;    }    return 0;}

【代码：AC】

#include <iostream>#include <iomanip>#include <cstring>#include <cstdlib>#include <cstdio>using namespace std;#define MAX 100+10struct Milk{    char name[MAX];    int price;    int volumn;};int cmp(const void *a, const void *b){    struct Milk *c = (Milk*)a;    struct Milk *d = (Milk*)b;    int day1 = c->volumn/200;    if (day1 >5)        day1 = 5;    int day2 = d->volumn/200;    if (day2 >5)        day2 = 5;    if ((c->price/day1) == (d->price/day2))        return c->volumn - d->volumn;    else        return (d->price/day2) - (c->price/day1);}int main(){    int N = 0;    cin >> N;    while (N--)    {        Milk milk[MAX];        int n = 0, i = 0;        cin >> n;        for (i = 0; i < n; )        {            Milk temp;            cin >> temp.name >> temp.price >> temp.volumn;            if (temp.volumn >= 200)                milk[i++] = temp;            else                n--;        }        qsort(milk, n, sizeof(milk[i]), cmp);        cout << milk[n-1].name << endl;    }    return 0;}

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