POJ2229 Sumsets 【递归】-优快云博客

本文介绍了一种解决特定类型组合数学问题的高效算法——求和集计数问题。问题要求找出所有可能的数集，这些数集由2的幂次组成且总和等于给定的整数N。算法采用动态规划的方法，通过递推公式逐步建立从1到N的解空间，并最终输出N的所有可能表示方式的最后九位数字。

Sumsets

Time Limit: 2000MS		Memory Limit: 200000K
Total Submissions: 13210		Accepted: 5300

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

Sample Output

Source

USACO 2005 January Silver

做这题时有些被曾经的经验束缚了，看完题第一反应是母函数，然后上模板。然后输入1000000等结果，等啊等就是不出结果。。

。參考了大牛的解题报告：Click

#include <stdio.h>
#include <string.h>

#define maxn 1000002
int dp[maxn];

int main() {
    int i, n;
    scanf("%d", &n);
    dp[1] = 1;
    for(i = 2; i <= n; ++i) {
        if(i & 1) dp[i] = dp[i-1];
        else dp[i] = dp[i-1] + dp[i/2];
        dp[i] %= 1000000000;
    }
    printf("%d\n", dp[n]);
    return 0;
}

转载于:https://www.cnblogs.com/hrhguanli/p/4828424.html