NYOJ 411 Friends number (数论--因子和)

输入

There are several cases.
Each test case contains two positive integers a, b(1<= a <= b <=5,000,000).
Proceed to the end of file.

输出

For each test case, output the number of couples in the given range. The output of one test case occupied exactly one line.

样例输入

1 100
1 1000

样例输出

0
1

提示

6 is a number whose sum of all divisors is 6. 6 is not a friend number, these number is called Perfect Number.

来源

辽宁省10年省赛

oj数据有点大，常规方法，一个劲的超时，看了一下排在前几名的人的代码，基本都是打表过，（囧~~），后来有点提示，用数组模拟过了，其实还可以用容器

，那样写的话看的比较舒服。

代码：（红色区域重点）

 
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
using namespace std;
#define N 5000001
#define CLR(arr, what) memset(arr, what, sizeof(arr))
int a[N] ,pre[N],last[N],ss=0;
void getsum();
int main()
{
    //getsum();
//    for(int i = 2; i <= N; i++) a[i] = 1;
//    for(int i = 2; i * i <= N; i++)
//        for (int j = i; j * i <= N; j++)
//            a[i * j] += i + j;
   <span style="color:#ff0000;"> for(int i = 2; i <= N; i++) a[i] = 1;
    for(int i = 2; i * i <= N; i++)
        for(int j = i + 1; i * j <= N; j++)
            a[i * j] += i + j;
    for(int i = 2; i * i <= N; i++)
        a[i * i] += i;
    for(int i=1; i<=N; i++)
    {
        int t=a[i];
        if (t > i && t <=N&& a[t] == i)
        {
            pre[ss]=i;
            last[ss]=t;
            ss++;
        }
    }</span>
    int m,i,j,n;
    while(~scanf("%d%d",&m,&n))
    {
        int count = 0;
        for(i=0; i<ss; i++)
            if(pre[i]>=m&&last[i]<=n)
                count++;
        printf("%d\n",count);
    }
    return 0;
}        

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