acm 1085母函数

输入1分，2分，5分的硬币，输出不能组成的最小总值；

#include"iostream"
using namespace  std;
int c1[10000],c2[10000];
int num[4];
int main()
{
   int nNum;
   while(scanf("%d %d %d",&num[1],&num[2],&num[3])&&(num[1]||num[2]||num[3]))
   {
        int _max=num[1]*1+num[2]*2+num[3]*5;
        for(int i=0;i<=_max;++i)
        {
           c1[i]=0;
           c2[i]=0;       
        }            
        for(int i=0;i<=num[1];++i)
          c1[i]=1;
        for(int i=0;i<=num[1];++i)
           for(int j=0;j<=num[2]*2;j+=2)
             c2[j+i]+=c1[i];
        for(int i=0;i<=num[2]*2+num[1]*1;++i)
        {
           c1[i]=c2[i];       
           c2[i]=0;
        }
        for(int i=0;i<=num[1]*1+num[2]*2;++i)
          for(int j=0;j<=num[3]*5;j+=5)
             c2[j+i]+=c1[i];
        for(int i=0;i<=num[2]*2+num[1]*1+num[3]*5;++i)
        {
           c1[i]=c2[i];
           c2[i]=0;       
        }
        int i;
        for(i=0;i<=_max;++i)
           if(c1[i]==0)
           {
              printf("%d\n",i);           
               break;
           }
        if(i==_max+1)
          printf("%d\n",i);
   }   
    return 0;
}

转载于:https://www.cnblogs.com/QQbai/archive/2011/08/05/2128330.html