hdu 1398 Square Coins/hdu 1028 Ignatius and the Princess III

两道母函数的模板题：

http://acm.hdu.edu.cn/showproblem.php?pid=1398

View Code

#include<iostream>#include<cstdio>using namespace std;int c1[310],c2[310];int main(){int n,i,j,k;while(scanf("%d",&n),n)    {for(i=0;i<=n;i++)        {            c1[i]=1;            c2[i]=0;        }for(i=2;i*i<=n;i++)        {for(j=0;j<=n;j++)for(k=0;k+j<=n;k+=i*i)        c2[k+j]+=c1[j];for(j=0;j<=n;j++)        {            c1[j]=c2[j];            c2[j]=0;        }        }        printf("%d\n",c1[n]);    }return 0;}

http://acm.hdu.edu.cn/showproblem.php?pid=1028

View Code

#include<iostream>#include<cstdio>using namespace std;int c1[130],c2[130];int main(){int n,i,j,k;while(scanf("%d",&n)!=EOF)    {for(i=0;i<=n;i++)        {            c1[i]=1;            c2[i]=0;        }for(i=2;i<=n;i++)        {for(j=0;j<=n;j++)for(k=0;k+j<=n;k+=i)        c2[k+j]+=c1[j];for(j=0;j<=n;j++)        {            c1[j]=c2[j];            c2[j]=0;        }        }        printf("%d\n",c1[n]);    }return 0;}

转载于:https://www.cnblogs.com/LT-blogs/archive/2011/11/26/2264499.html

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