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Poj2229--Sumsets(递推)

时间:2015-12-21 20:17:34      阅读:186      评论:0      收藏:0      [点我收藏+]

标签:

Sumsets
Time Limit: 2000MS   Memory Limit: 200000K
Total Submissions: 15052   Accepted: 6001

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

Source

 
题目很有意思, 没想到是个递推, 脑子不够用。
#include <cmath>
#include <cstdio> 
const int  MOD = 1e9;
const int N = 1000001;
int num[N];
void Sieve(){
    num[1] = 1;
    num[2] = 2;
    num[3] = 2;
    for(int i = 4; i <= N; i++){
        if(i & 1)
            num[i] = num[i-1];
        else
            num[i] = (num[i-2]%MOD + num[i/2]%MOD)%MOD;
    }
}
int main(){
    Sieve();
    int n;
    while(scanf("%d", &n) != EOF)
        printf("%d\n", num[n]);
    return 0;
}

 

Poj2229--Sumsets(递推)

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原文地址:http://www.cnblogs.com/fengshun/p/5064645.html

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