NBUT 1223 Friends number

最新推荐文章于 2019-04-01 11:38:51 发布

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本文深入探讨了信息技术领域的核心概念及其在实际应用中的重要性，涵盖了从开发工具到人工智能等多个细分领域，提供了全面的技术概览。

[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

题目意思是说，若一对正整数A和B，假设A的所有真因子（不含本身）之和为sumA，B的所有真因子（不含本身）之和为sumB，满足sumA=B ，sumB=A，这样的称为友谊数。输入给定的区间，输出区间内有多少对友谊数。

分析：数据有500W，非常大，直接做是不可能的，那唯一的办法就是打表了。

打表代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
const int mx=5000000+10;
int ans[mx];
int main() {
    //freopen("out.txt","w",stdout);
    ans[0] = ans[1] = 1;
    memset(ans,0,sizeof(ans));
    for(int i = 2; i <= mx; i++){
        ans[i]++;
        for(int j = 2*i; j <= mx; j += i){
            ans[j] += i;
        }
    }
    int cnt=0;
    for(int i = 2; i <= mx; i++){
        if(ans[i] <= mx && i == ans[ans[i]] && i < ans[i]){
            printf(" %d , %d , ",i,ans[i]);
            cnt++;
            if(cnt%4==0)
                puts("");
        }
    }
}

打表后就很简单了，AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
int a[145]= { 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(~scanf("%d%d",&x,&y))
    {
        int cnt=0;
        for(int i=0;i<145;i+=2)
        {
            if(a[i]>=x && y>=a[i+1])
            {
                //printf("%d %d\n",a[i],a[i+1]);
                cnt++;
            }
        }
        printf("%d\n",cnt);
    }
    return 0;
}