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  • POJ2229 Sumsets 【递归】

    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

    7

    Sample Output

    6

    Source

    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
    ", dp[n]);
        return 0;
    }


    版权声明:本文博主原创文章。博客,未经同意不得转载。

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  • 原文地址:https://www.cnblogs.com/hrhguanli/p/4828424.html
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