九度 oj 题目1084：整数拆分

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

照抄了 http://blog.csdn.net/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]);     }  }

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