Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12639 Accepted Submission(s): 8929
4 10 20
5 42 627
#include<stdio.h>
int main()
{
int n,i,j,k;
int a[1010],b[1010];//这个根据题目要求灵活变化
while(~scanf("%d",&n))//要是有对n的其他限制条件也可以在此 说明 加以限制
{
for(i=0;i<=n;i++)
a[i]=1,b[i]=0;
for(i=2;i<=n;i++)
{
for(j=0;j<=n;j++)
for(k=0;k+j<=n;k+=i)
b[j+k]+=a[j];
for(j=0;j<=n;j++)
a[j]=b[j],b[j]=0;
}
printf("%d\n",a[n]);
}
return 0;
}<span style="font-size:14px;">#include<stdio.h>
int main()
{
int n;
int i,j,k;
int a[130],b[130];
for(i=0;i<=129;i++)
a[i]=1,b[i]=1;
for(i=2;i<=129;i++)
{
for(j=1;j<=129/i;j++)
{
for(k=0;k<=129;k++)
if(k+j*i<=129)
b[k+j*i]+=a[k];
}
for(j=0;j<=129;j++)
a[j]=b[j];
}
while(~scanf("%d",&n))
{
printf("%d\n",a[n]);
}
return 0;
}</span>原文地址:http://blog.csdn.net/ice_alone/article/details/38370713