Holding Bin-Laden Captive!_hdu_1085(DP).java

/*
 * 9607741 2013-11-17 18:04:23 Accepted 1085 187MS 5700K 1251 B Java zhangyi
 http://acm.hdu.edu.cn/showproblem.php?pid=1085
 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12778    Accepted Submission(s): 5728


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
 
 */


/*
 * 大致题意：给你面值为1,2,5的硬币，不能组成最小的面值为？
 * 

 */

 

import java.io.InputStreamReader;
import java.util.Scanner;


public class Main {//DP
	public static void main(String[] args) {
		Scanner input=new Scanner(new InputStreamReader(System.in));
		while(true){
			int a=input.nextInt();
			int b=input.nextInt();
			int c=input.nextInt();
			if(a+b+c==0)
				break;
			Mon m[]=new Mon[a+b*2+c*5+1];
			for(int i=1;i<=a+b*2+c*5;i++)
				m[i]=new Mon();
			m[0]=new Mon(a,b,c);
			m[0].ok=true;
			for(int i=0;i<=a+b*2+c*5;i++){
				if(m[i].ok){
					if(m[i].a>0){
						m[i+1].a=m[i].a-1;
						m[i+1].b=m[i].b;
						m[i+1].c=m[i].c;
						m[i+1].ok=true;
					}
					if(m[i].b>0){
						m[i+2].a=m[i].a;
						m[i+2].b=m[i].b-1;
						m[i+2].c=m[i].c;
						m[i+2].ok=true;
					}
					if(m[i].c>0){
						m[i+5].a=m[i].a;
						m[i+5].b=m[i].b;
						m[i+5].c=m[i].c-1;
						m[i+5].ok=true;
					}
				}
			}
			boolean okk=true;
			for(int i=1;i<=a+b*2+c*5;i++){
				if(!m[i].ok)
				{
					System.out.println(i);
					okk=false;
					break;
				}
			}
			if(okk)
				System.out.println(a+b*2+c*5+1);
		}
	}
}
class Mon{
	boolean ok=false;
	int a,b,c;
	public Mon(int a, int b, int c) {
		this.a = a;
		this.b = b;
		this.c = c;
	}
	Mon(){};
}