HDU2709Sumsets整数分解为加数，递推公式

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=2709

Sumsets

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2830    Accepted Submission(s): 1134

Problem 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

7

 

Sample Output

6

题解：这个题当数据量不大的时候，完全可以用搜索得出结果，但数据量大时，会超时。所以，可以先用dfs搜索得出前面若干项的结果，看一下有什么规律。

这个题从1开始，结果一次为 1   2   2   4   4   6   6   10   10   14   14   20   20   26   26   36   36   46   46   60   60   74。

递推公式：p[ i ]=(p[ i-1 ]+p[ i/2 ]) % mod ; （mod=1e9）

AC代码：

#include<stdio.h>typedef long long ll;ll mod=1000000000;ll p[1020000];void init(){    p[1]=1;p[2]=2;    for(int i=2;i<1000010;i+=2)    {        p[i]=(p[i-1]+p[i/2])%mod;        p[i+1]=p[i];    }}int main(){    int n;    init();    while(~scanf("%d",&n))    {        printf("%lld\n",p[n]);    }    return 0;}

dfs搜索过程：（非AC代码）

#include<stdio.h> int pow2[100]={1,2,4,8,16,32,64,128,256,512,1024,2048,4096};int ans;void dfs(int n,int last){    if(n==0)    {        ans++;        return;    }    for(int i=0;n-pow2[i]>=0;i++)    {        if(pow2[i]>=last)            dfs(n-pow2[i],pow2[i]);    }}int main(){    int n;    //while(~scanf("%d",&n))    for(n=1;n<40;n++)    {        ans=0;        dfs(n,0);        printf("   %d",ans);    }    return 0;}