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

完全背包..

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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).

给出一个N(1≤N≤10^6)，使用一些2的若干次幂的数相加来求之．问有多少种方法

Input

一个整数N.

Output

方法数．这个数可能很大，请输出其在十进制下的最后9位．

Sample Input

7

Sample Output

6

有以下六种方式
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

HINT

Source

Silver