ZOJ Problem Set - 2095 Divisor Summation

Divisor Summation

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Time Limit: 5 Seconds      Memory Limit: 32768 KB

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Give a natural number n (1 <= n <= 500000), please tell the summation of all its proper divisors.

Definition: A proper divisor of a natural number is the divisor that is strictly less than the number.

e.g. number 20 has 5 proper divisors: 1, 2, 4, 5, 10, and the divisor summation is: 1 + 2 + 4 + 5 + 10 = 22.

Input

An integer stating the number of test cases, and that many lines follow each containing one integer between 1 and 500000.

Output

One integer each line: the divisor summation of the integer given respectively.

Sample Input

3
2
10
20

Sample Output

1
8
22

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Author: Neal Zane
Source: ZOJ Monthly, March 2004

分析：

题意：

定义：一个自然数的严格意义上的除数是比它本身小的除数。

给定一个自然数n（1<=n<=500000），要求输出它的所有严格意义上的除数的和。

先打表预处理所有500000以内的自然数的除数和，然后根据输入直接查询就行。这里的打表有一定技巧。

ac代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=500000+10;
int a[maxn];


void table()//打表，注意技巧
{
    int i,j;
    int m=sqrt(500000);
    //memset(a,1,sizeof(a));//memset置1出问题
    for(i=2;i<=500000;i++)
    a[i]=1;
    a[1]=0;
    for(i=2;i<=m;i++)//比如64的除数有1,2,4,8,16,32.则64=2*32=4*16=8*8，通过两重for循环可全部得到
    {
        a[i*i]+=i;//得到8
        for(j=i+1;j<=(500000/i);j++)
        {
            a[i*j]+=i+j;//分别得到2*32,4*16
        }
    }
}
int main()
{
    int t,n;
    table();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        printf("%d\n",a[n]);
    }
    return 0;
}