NBUT 1223 Friends number

Paula and Tai are couple. There are many stories between them. The day Paula left by airplane, Tai send one message to telephone [phone], 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]。

Input

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.

Output

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

Sample Input

1 1001 1000

Sample Output

01

Hint

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

输出所有约数和==别外一个数  别外一个数的约数和等于自己的对数 答案说到500W 结果只要50W就水过去了

/*#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cstdlib>using namespace std;int a[5000005];int main(){int i,j;memset(a,0,sizeof a);for(i=1;i<=500000;i++){for(j=1;i*j<=500000;j++){a[i*j]+=i;}}for(i=1;i<=500000;i++)a[i]-=i;int k=0;for(i=1;i<=500000;i++){if(i==a[a[i]] && i!=a[i]){if(i>a[i])continue;printf("%d,",a[i]);k++;}}return 0;}*/#include <stdio.h>int a[]={220,284,1184,1210,2620,2924,5020,5564,6232,6368,10744,10856,12285,14595,17296,18416,63020,76084,66928,66992,67095,71145,69615,87633,79750,88730,100485,124155,122265,139815,122368,123152,141664,153176,142310,168730,171856,176336,176272,180848,185368,203432,196724,202444,280540,365084,308620,389924,319550,430402,356408,399592,437456,455344,469028,486178};int main(){    int c,d;    while(~scanf("%d %d",&c,&d))    {        int ans=0;        for(int i=0;i<56;i+=2)        {            if(c<=a[i] && a[i+1]<=d)                ans++;        }        printf("%d\n",ans);    }    return 0;}