Friends number NBUT - 1223 （暴力打表）

Paula and Tai are couple. There are many stories between them. The day Paula left by airplane, Tai send one message to telephone 2200284, then, everything is changing… (The story in “the snow queen”).

After a long time, Tai tells Paula, the number 220 and 284 is a couple of friends number, as they are special, all divisors of 220’s sum is 284, and all divisors of 284’s sum is 220. Can you find out there are how many couples of friends number less than 10,000. Then, how about 100,000, 200,000 and so on.

The task for you is to find out there are how many couples of friends number in given closed interval [a,b]。

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.

//对5000000之内的每个数的所有因子暴力打表求和竟然不会超时，我擦。。。很强

#include<map>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#include<cstdio>
#include<sstream>
#include<numeric>//STL数值算法头文件
#include<stdlib.h>
#include <ctype.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<functional>//模板类头文件
using namespace std;

typedef long long ll;
const int maxn=11000;
const int INF=0x3f3f3f3f;

int f[5000010];
int xA[200],xB[200];

int main()
{
    for(int i=1; i<=5000000; i++)//暴力打表,把每个数的所以因子的和求出来
    {
        int xsk=i+i;
        while(xsk<=5000000)
        {
            f[xsk]+=i;//f函数存储的是每个数所以因子的和
            xsk+=i;
        }
    }

    int sum=0;
    for(int i=1; i<=5000000; i++)
    {
        int A=i,B=f[i];
        if(B>A&&B<5000000&&f[B]==i)
        {
            xA[sum]=A;
            xB[sum]=B;
            sum++;
        }
    }

    int L,R;
    while(~scanf("%d%d",&L,&R))
    {
        int ans=0;
        for(int i=0; i<sum; i++)
        {
            if(xA[i]>=L&&xA[i]<=R&&xB[i]>=L&&xB[i]<=R) ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}