母函数练习2 HDU 1085-优快云博客

本文探讨了恐怖分子本·拉登在中国杭州隐匿期间，因长时间无聊而投身于数学问题的故事。故事中，本·拉登设定了一道数学题：给定三种不同面额的硬币（1元、2元、5元），分别有num_1、num_2、num_5个，求不能用这些硬币支付的最小正整数值。作为超级ACMer，解决这个问题轻而易举，并有机会从布什那里获得2500万美元。

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Problem Description

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

#include <stdio.h>
#include <string.h>
int num1,num2,num3,arr[10001],tmp[10001];

int main(){
	freopen("in.txt","r",stdin);
	while(true){
		scanf("%d %d %d", &num1, &num2, &num3);
		if(num1 == 0 && num2 == 0 && num3 == 0)
			break;
	//	int l = num1 + num2 * 2 + num3 * 5;
		int i,j;
		memset(arr,0,sizeof(arr));
		memset(tmp,0,sizeof(tmp));
		for(i=0; i<=num1; i++){
			arr[i] = 1;
			tmp[i] = 0;
		}
	//	int m = max(num1, num2 * 2 + num1);
		int m = num2 * 2 + num1;
	//	printf("%d\n",m);
		for(i=0; i<=m; i++){
			for(j=0; j+i <= m; j += 2){
				tmp[j+i] += arr[i];
			}
		}
		for(i=0; i<=m; i++){
			arr[i] = tmp[i];
			tmp[i] = 0;
		}

		m += 5*num3;

		for(i=0; i<=m; i++){
			for(j=0; j+i <= m; j += 5){
				tmp[j+i] += arr[i];
			}
		}
		
		for(i=1; i<=m; i++){
			if(tmp[i] == 0)
				break;
		}
		printf("%d\n",i);

	}
	return 0;
}