hdu1085：Holding Bin-Laden Captive!

http://acm.hdu.edu.cn/showproblem.php?pid=1085

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

题意分析：

三种硬币面值分别为1元，2元，5元，给出三种硬币的个数，问不能组成的最小钱数为多少。

解题思路：

只要1元和2元的可以组成4，就可以组成所有硬币面值之和以下的所有数字，否则就只能组合1元和2元硬币面值之和以下的所有数字。

#include <stdio.h>
int main()
{
    int a, b, c;
    while(scanf("%d%d%d", &a, &b, &c)!=EOF)
    {
        if(a==0 && b==0 && c==0)
            break;
        if(a==0)
            printf("1
");
        else if(a+b*2<4)
            printf("%d
", a+b*2+1);
        else
            printf("%d
", a+b*2+c*5+1);
    }
    return 0;
}