hdoj 1085 Holding Bin-Laden Captive! <思维>

最新推荐文章于 2019-06-29 14:22:25 发布

leibniz_zhang

最新推荐文章于 2019-06-29 14:22:25 发布

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本·拉登藏匿于中国杭州的一个洞穴中，并提出一个数学难题：给出三种硬币（1元、2元、5元），数量分别为num_1、num_2和num_5，请输出无法用这些硬币支付的最小正数值。本文提供了一个解决该问题的算法。

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Holding Bin-Laden Captive!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20230 Accepted Submission(s): 9015

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

Author

lcy

代码：

#include<cstdio>
#define MA 9000
bool fafe[MA];
int shu[MA],ge;
int main()
{
    int a,b,c;
    while (scanf("%d%d%d",&a,&b,&c),a+b+c)
    {
        if (a==0)
        {
            printf("1\n");
            continue;
        }
        int kp=a+2*b+1;
        for (int i=0;i<kp;i++)
            shu[i]=1;
        if (kp<5)
        {
            printf("%d\n",kp);
        }
        else
            printf("%d\n",kp+c*5);
    }
    return 0;
}