九度 oj 题目1084：整数拆分

http://ac.jobdu.com/problem.php?pid=1084

照抄了 http://blog.youkuaiyun.com/jdplus/article/details/18839343

参考了 http://www.cnblogs.com/yinger/archive/2012/08/17/2644312.html

#include <stdio.h>

int main(){ 
    int a[1000002]; 
    int input;
    a[0] = a[1] = 1;
    for (int i = 1; i <=500000; ++i) { 
        a[2*i] = (a[2*i-2] + a[i])%1000000000; 
        a[2*i+1] = a[2*i];
    } 

    while(scanf("%d",&input) !=EOF ){ 
        printf("%d\n",a[input]);  
    }  
}  

这道题，如果找不到递推公式，也可以用完全背包的动态规划来做

#include <stdio.h>
#include <cstring>
static int dp[1000001];


int main(){ 
    int n; 
    while(scanf("%d",&n)!=EOF){ 
       memset(dp,0,sizeof(dp));
       dp[0] =1;
       for (int i = 0; i < 21; ++i) { 
           int v = 1<<i;
           for (int j = 1;  j<=n; ++j) { 
                if(j-v>=0){ 
                    dp[j] = (dp[j-v] + dp[j])%1000000000; 
                }       
           }  
       }  
       printf("%d\n",dp[n]); 
    }  
}