poj 2229
Sumsets
Time Limit: 2000MS Memory Limit: 200000K Total Submissions: 6980 Accepted: 2763 Description
Farmer John commanded his cows to search for different sets of numbers that sum to a given number. The cows use only numbers that are an integer power of 2. Here are the possible sets of numbers that sum to 7:
1) 1+1+1+1+1+1+1
2) 1+1+1+1+1+2
3) 1+1+1+2+2
4) 1+1+1+4
5) 1+2+2+2
6) 1+2+4
Help FJ count all possible representations for a given integer N (1 <= N <= 1,000,000).Input
A single line with a single integer, N.Output
The number of ways to represent N as the indicated sum. Due to the potential huge size of this number, print only last 9 digits (in base 10 representation).Sample Input
7Sample Output
6Source
难得写次递归,vs2008居然爆掉了》Stack overflow》,郁闷了下
据说是栈溢出,不清楚是什么情况
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
int f[1000002];
int find(int num)
{
if(f[num])
return f[num];
if(num==0)
return 0;
if(num%2==0)
f[num]=(find(num-2)+find(num/2))%1000000000;
else if(num%2==1)
f[num]=find(num-1)%1000000000;
return f[num];
}
int main()
{
int n;
memset(f,0,sizeof(f));
f[1]=1;
f[2]=2;
while(scanf("%d",&n)!=EOF)
{
printf("%d\n",find(n));
}
return 0;
}
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原文来自博客园——Submarinex的博客: www.cnblogs.com/submarinex/               
 
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