UVA 10465 Homer Simpson(dp + 完全背包)

Problem C:	Homer Simpson
Time Limit: 3 seconds Memory Limit: 32 MB

Homer Simpson, a very smart guy, likes eating Krusty-burgers. It takes Homer m minutes to eat a Krusty- burger. However, there�s a new type of burger in Apu�s Kwik-e-Mart. Homer likes those too. It takes him n minutes to eat one of these burgers. Given t minutes, you have to find out the maximum number of burgers Homer can eat without wasting any time. If he must waste time, he can have beer.

Input

Input consists of several test cases. Each test case consists of three integers m, n, t (0 < m,n,t < 10000). Input is terminated by EOF.

Output

For each test case, print in a single line the maximum number of burgers Homer can eat without having beer. If homer must have beer, then also print the time he gets for drinking, separated by a single space. It is preferable that Homer drinks as little beer as possible.

Sample Input

3 5 54

3 5 55

Sample Output

18

17

题意：有两种汉堡一种吃一个要n分钟，一种要m分钟，现在给定t分钟。要求尽量不浪费时间去吃汉堡，求出吃汉堡的个数，如果一定要浪费时间，那么还要输出浪费掉的时间。

思路：完全背包，等于是给定两个物品，时间代表容量，价值都为1，求出dp[t]的最大值。

代码：

#include <stdio.h>

#include <string.h>

int m, n, t, dp[10005], i;

int max(int a, int b) {

	return a > b ? a : b;

}

int main() {

	while (~scanf("%d%d%d", &m, &n, &t)) {

		memset(dp, -1, sizeof(dp));

		dp[0] = 0;

		for (i = m; i <= t; i ++) {

			if (dp[i - m] != - 1)

				dp[i] = max(dp[i - m] + 1, dp[i]);

		}

		for (i = n; i <= t; i ++) {

			if (dp[i - n] != -1)

				dp[i] = max(dp[i - n] + 1, dp[i]);

		}

		if (dp[t] != -1)

			printf("%d", dp[t]);

		else {

			for (i = t; i >= 0; i --)

				if (dp[i] != -1) {

					printf("%d %d", dp[i], t - i);

					break;

				}

		}

		printf("\n");

	}

	return 0;

}

巴特西

UVA 10465 Homer Simpson(dp + 完全背包)

Input

Output

Sample Input

Sample Output

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