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Holding Bin-Laden Captive!

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!

1 1 3
0 0 0

4

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#define N 10009
using namespace std;

int a[N];
int c1[N],c2[N];
int b[3]={1,2,5};

int main()
{
    while(~scanf("%d %d %d",&a[0],&a[1],&a[2]))
    {
        if(a[1]+a[2]+a[0]==0) break;

        int sum=0;
        for(int i=0;i<3;i++)
         sum+=a[i]*b[i];

         memset(c1,0,sizeof c1);
         memset(c2,0,sizeof c2);

         for(int i=0;i<=a[0];i++)//只有1面值的情况
         c1[i]=1;

        for(int i=1;i<3;i++)
        {
            for(int j=0;j<=sum;j++)
            if(c1[j])
            {
                for(int k=0;j+k<=sum&&k<=a[i]*b[i];k+=b[i])
                c2[j+k]+=c1[j];
            }

            for(int j=0;j<=sum;j++)
                c1[j]=c2[j],c2[j]=0;
        }

        int ans=0;
        for(int i=1;i<=10000;i++)
        if(c1[i]==0)
         {ans=i;break;}

         cout<<ans<<endl;
    }
    return 0;
}

HDU 1085 简单母函数

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原文地址：http://blog.csdn.net/wust_zjx/article/details/44259735