NBUT1223 Friends number（打表，暴力）

题目：

• [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 [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]。

  
  

  
  

  
  

  
  

  
  

  
  

  
  

  
  

• 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 1001 1000

• 样例输出

• 01

• 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

• 操作
•     



思路：

这个题的题意我看了好久才看懂，果然还是太菜了。。

题目定义了一个【友谊数】，它的定义是：两个数a和b，a的所有因子（不包含本身）的和加起来等于b，b的所有因子（不包含本身）的和加起来恰好等于a，这样的数被称为友谊数,数据量有50万，先打表处理，再解决。

打表代码：

#include <cstdio>#include <cstring>#include <cctype>#include <string>#include <set>#include <iostream>#include <stack>#include <cmath>#include <queue>#include <vector>#include <algorithm>#define mem(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3f#define N 5000000+20#define mod 10007#define M 1000000+10#define LL long longusing namespace std;int num[N];//它的因子数之和是多少int main(){    num[0]=num[1]=1;    for(int i=2;i<=N;i++)    {        num[i]++;        for(int j=2*i;j<=N;j+=i)        {            num[j]+=i;        }    }    for(int a=2;a<=N;a++)    {        int b=num[a];//b等于a的所有因子数之和        if(b<=N&&a==num[b]&&a<b)//b在N的范围内，a等于b的所有因子数之和，a<b            printf("%d,%d,",a,b);    }    return 0;}

解题代码：

#include <cstdio>#include <cstring>#include <cctype>#include <string>#include <set>#include <iostream>#include <stack>#include <cmath>#include <queue>#include <vector>#include <algorithm>#define mem(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3f#define N 5000000+20#define mod 10007#define M 1000000+10#define LL long longusing namespace std;int a[10000]={    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 x,y;    while(cin>>x>>y)    {        int sum=0;        for(int i=0; i<=150; i+=2)        {            if(a[i]>=x&&a[i+1]<=y)                sum++;        }        cout<<sum<<endl;    }    return 0;}