Friends number

最新推荐文章于 2018-08-03 08:48:09 发布

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最新推荐文章于 2018-08-03 08:48:09 发布

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分类专栏： acm算法之路

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本文探讨了友好数的概念及如何找出特定区间内的友好数对。通过预先计算并存储1到500万之间的所有友好数对，文章提供了一种高效的方法来解决在给定区间内查找友好数对数量的问题。

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Friends number

时间限制: 1 Sec 内存限制: 128 MB

题目描述

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.

题意概括

如果a的所有因子和为b，而且b的所有因子和为a，那么就成a和b为友好数字，给一个区间，问在这个区间内有多少队友好数字。

解题思路

将1到5000000以内的所有友好数字打表，之后判断给定区间内一共有多少个，为了防止超时，我们将所有友好数字输出，然后初始化到一个二维数组里面。

代码

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<ctype.h>
#include<algorithm>
using namespace std;

int line[1000][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
,4482765,5120595
,4532710,6135962
,4604776,5162744
,5123090,5504110
,5147032,5843048
,5232010,5799542
,5357625,5684679
,5385310,5812130
,5459176,5495264
,5726072,6369928
,5730615,6088905
,5864660,7489324
,6329416,6371384
,6377175,6680025
,6955216,7418864
,6993610,7158710};
int main()
{
    int i,j,a,b,sum;
    while(~scanf("%d %d",&a,&b)){
        sum=0;
        for(i=0;i<84;i++){
            if(line[i][0]>=a&&line[i][1]<=b){
                sum++;
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}