# BZOJ 1677: [Usaco2005 Jan]Sumsets 求和( dp )

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#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>

#define rep( i , n ) for( int i = 0 ; i < n ; i++ )
#define clr( x , c ) memset( x , c , sizeof( x ) )

using namespace std;

#define mod( x )  ( ( x ) %= 1000000000 )

const int maxn = 1000000 + 5;

int d[ maxn ];

int main() {
freopen( "test.in" , "r" , stdin );
int n;
cin >> n;
clr( d , 0 );
d[ 0 ] = 1;
for( int i = 0 ; ( 1 << i ) <= n ; i++ )
for( int j = 1 << i ; j <= n ; j++ )
mod( d[ j ] += d[ j - ( 1 << i ) ] );

cout << d[ n ] << "\n";
return 0;
}

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## 1677: [Usaco2005 Jan]Sumsets 求和

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 666  Solved: 367
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## 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).

一个整数N.

7

6

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

## Source

posted @ 2015-06-09 13:56  JSZX11556  阅读(176)  评论(0编辑  收藏  举报