fzuoj1607Greedy division

时间：2014-07-06 00:44:58 阅读：458 评论：0 收藏：0 [点我收藏+]

题目链接：

题目：

Problem 1607 Greedy division

Accept: 436 Submit: 1590
Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem Description

Oaiei has inherited a large sum of wealth recently; this treasure has n pieces of golden coins. Unfortunately, oaiei can not own this wealth alone, and he must divide this wealth into m parts (m>1). He can only get one of the m parts and the m parts have equal coins. Oaiei is not happy for only getting one part of the wealth. He would like to know there are how many ways he can divide the wealth into m parts and he wants as many golden coins as possible. This is your question, can you help him?

Input

There are multiply tests, and that will be 500000. For each test, the first line is a positive integer N(2 <= N <= 1000000), N indicates the total number of golden coins in the wealth.

Output

For each test, you should output two integer X and Y, X is the number of ways he can divide the wealth into m parts where m is large than one, Y is the number of golden coins that he can get from the wealth.

Sample Input

568

Sample Output

1 13 33 4

Hint

There are huge tests, so you should refine your algorithm.

Source

FOJ月赛-2008年5月

思路：

这个题目首先预处理，其实这个题目就是在求有多少个因数，所以可以用素数打表的思路，因为每次扫描到的j，一定是i的倍数，故a[j]++，例如是24，则i到2，3，4，6，8，12，24时都会加一次。。。然后就是如果是素数，则直接输出1和1.相当于一种优化吧。。

代码为：

#include<cstdio>
#include<cstring>
const int maxn=1000000+10;
int a[maxn];
int main()
{
    int n,i;
    memset(a,0,sizeof(a));
    for(int i=2;i<=maxn;i++)
        for(int j=i;j<=maxn;j=j+i)
            a[j]++;
    while(~scanf("%d",&n))
    {
        if(a[n]==1)
            printf("%d %d\n",1,1);
        else
            {
                for(i=2;i<=maxn/2;i++)
                    if(n%i==0)
                     break;
                printf("%d %d\n",a[n],n/i);
            }

    }
    return 0;
}

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fzuoj1607Greedy division

原文：http://blog.csdn.net/u014303647/article/details/36736549

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