杭电1085（多重背包求解）

题目：

We all know that Bin-Laden is a notorious terrorist, and he has disappeared for a long time. But recently, it is reported that he hides in Hang Zhou of China!
“Oh, God! How terrible! ”

Don’t be so afraid, guys. Although he hides in a cave of Hang Zhou, he dares not to go out. Laden is so bored recent years that he fling himself into some math problems, and he said that if anyone can solve his problem, he will give himself up!
Ha-ha! Obviously, Laden is too proud of his intelligence! But, what is his problem?
“Given some Chinese Coins (硬币) (three kinds-- 1, 2, 5), and their number is num_1, num_2 and num_5 respectively, please output the minimum value that you cannot pay with given coins.”
You, super ACMer, should solve the problem easily, and don’t forget to take $25000000 from Bush!

 

Input

Input contains multiple test cases. Each test case contains 3 positive integers num_1, num_2 and num_5 (0<=num_i<=1000). A test case containing 0 0 0 terminates the input and this test case is not to be processed.

 

Output

Output the minimum positive value that one cannot pay with given coins, one line for one case.

 

Sample Input

1 1 3

0 0 0

 

Sample Output

4

思想：

　　给你1 2 5价值的钱，数量有限，可以用多重背包问题求解。

算出总的钱数，当做背包容量。相当于给你1你可以买1价值的东西，

给你4你却只能买3价值的东西。所以显然到最后，如果dp[i] < i;说明

所给的钱数不可以购买。

 

代码：

 1 #include<stdio.h>

 2 #include<string.h>

 3 

 4 

 5 int getmax(int x, int y){

 6     return x > y ? x : y;

 7 }

 8 

 9 int main(){

10     int sumprice, price[3], num[3001], i, j, cout, dp[8001], k;

11     while(scanf("%d %d %d", &price[0], &price[1], &price[2]) && (price[0] || price[1] || price[2])){

12         cout = 0;

13         k = 0;

14         sumprice = price[0] + 2 * price[1] + 5 * price[2];

15         while(price[0] --){

16             num[cout ++] = 1;

17         }

18         while(price[1] --){

19             num[cout ++] = 2;

20         }

21         while(price[2] --){

22             num[cout ++] = 5;

23         }

24         memset(dp, 0, sizeof(dp));

25         for(i = 0; i < cout; i ++){

26             for(j = sumprice; j >= num[i]; j --){

27                 dp[j] = getmax(dp[j], dp[j - num[i]] + num[i]);

28             }

29         }

30         for(i = 1; i <= sumprice; i ++){

31             if(dp[i] < i){

32                 printf("%d\n", i);

33                 k = 1;

34                 break;

35             }

36         }

37         if(k == 0){

38             printf("%d\n", sumprice + 1);

39         }

40     }

41     return 0;

42 }

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