NBUT 1223 Friends number (打表)

[1223] Friends number

时间限制: 1000 ms　内存限制: 131072 K

问题描述

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.

来源

辽宁省赛2010

题意：如果A所有因子（不包括自身）的和等于B，并且，B所有因子（不包括自身）的和等于A，则称A，B互为friend number，求给定区间有有多少对friend number

分析：这道题一开始的想法是先生成所有数的friend number ，然后O(n)查询，但是发现500W的数据很大，会TLE，所以决定打表，打表写条件的时候要注意，如果得到的数字大于500W就舍弃。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<string>
#include<iostream>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<map>
#include<stdlib.h>
#include<algorithm>
#define LL __int64
#define FIN freopen("in.txt","r",stdin)
using namespace std;
int a[71][2]={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
,503056,514736
,522405,525915
,600392,669688
,609928,686072
,624184,691256
,635624,712216
,643336,652664
,667964,783556
,726104,796696
,802725,863835
,879712,901424
,898216,980984
,947835,1125765
,998104,1043096
,1077890,1099390
,1154450,1189150
,1156870,1292570
,1175265,1438983
,1185376,1286744
,1280565,1340235
,1328470,1483850
,1358595,1486845
,1392368,1464592
,1466150,1747930
,1468324,1749212
,1511930,1598470
,1669910,2062570
,1798875,1870245
,2082464,2090656
,2236570,2429030
,2652728,2941672
,2723792,2874064
,2728726,3077354
,2739704,2928136
,2802416,2947216
,2803580,3716164
,3276856,3721544
,3606850,3892670
,3786904,4300136
,3805264,4006736
,4238984,4314616
,4246130,4488910
,4259750,4445050
};
int main()
{
int l,r;
while(scanf("%d %d",&l,&r)!=EOF)
{
int ans=0;
for(int i=0;i<71;i++)
{
if(l<=a[i][0] && r>=a[i][1])
ans++;
}
printf("%d\n",ans);
}
return 0;
}

View Code