hdu 1085 Holding Bin-Laden Captive!

问题描述：“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.”

这是一道基本的母函数应用题。

#include<stdio.h>
#include<string.h>
int c1[8010],c2[8010];
int main()
{
    int a,b,c,sum,a1[4]={2,5},i,j,k;
    while(scanf("%d%d%d",&a,&b,&c)!=EOF)
    {
        if(a==0&&b==0&&c==0)
            break;
        sum=a+b*2+c*5;
        a1[2]=b;a1[3]=c;
        memset(c1,0,sizeof(c1));
        memset(c2,0,sizeof(c2));
        for(i=0;i<=a;i++)//初始化
            c1[i]=1;
        int len=a;
        for(i=2;i<=3;i++)//i表示第i个表达式
        {
            for(j=0;j<=len;j++)//j表示前面i-1个表达式相乘后得到的表达式的每一项的系数
                for(k=0;k<=a1[i]*a1[i-2];k+=a1[i-2])//k表示第i个表达式每一项的系数
                    c2[k+j]+=c1[j];//将系数进行合并
            len+=a1[i-2]*a1[i];
            for(j=0;j<=len;j++)
            {
                c1[j]=c2[j];
                c2[j]=0;
            }
        }
        for(i=1;i<=sum;i++)    
            if(c1[i]==0)
            {
                printf("%d
",i);
                break;    
            }
        if(i>sum) printf("%d
",sum+1);
    }
    return 0;
}