ACdream 1115 Salmon And Cat (找规律&&打表)-白红宇

ACdream 1115 Salmon And Cat (找规律&&打表)

阅读量：5051 次

发布时间：2019-06-12

本文共 2022 字，大约阅读时间需要 6 分钟。

题目链接：

题意：

一个数被觉得是一个完美的数，仅仅要须要满足下面的两个条件之中的一个

1)x = 1 or 3

2)x = 2 + a*b + 2*a + 2*b; a。b都是完美的数。

分析：

x + 2 = (a + 2)*(b + 2)

因为x1=1,x2=3。全部的数都是由着两个数衍生而来。那么我们就可

以得出一个结论了。一个数x假设是完美的数。那么x = 3^p*5^q;

因此代码例如以下：

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           using namespace std;int main(){ int n; while(~scanf("%d",&n)){ n+=2; while(n%3==0) n/=3; while(n%5==0) n/=5; if(n!=1) puts("No"); else puts("Yes"); } return 0;}

还能够打表把1e9以内的全部完美数打出来。

代码例如以下：

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           using namespace std;typedef long long LL;LL a[140]={1,3,7,13,23,25,43,73,79,123,133,223,241,373,403,623,673,727,1123,1213,1873,2023,2185,3123,3373,3643,5623,6073,6559,9373,10123,10933,15623,16873,18223,19681,28123,30373,32803,46873,50623,54673,59047,78123,84373,91123,98413,140623,151873,164023,177145,234373,253123,273373,295243,390623,421873,455623,492073,531439,703123,759373,820123,885733,1171873,1265623,1366873,1476223,1594321,1953123,2109373,2278123,2460373,2657203,3515623,3796873,4100623,4428673,4782967,5859373,6328123,6834373,7381123,7971613,9765623,10546873,11390623,12301873,13286023,14348905,17578123,18984373,20503123,22143373,23914843,29296873,31640623,34171873,36905623,39858073,43046719,48828123,52734373,56953123,61509373,66430123,71744533,87890623,94921873,102515623,110716873,119574223,129140161,146484373,158203123,170859373,184528123,199290373,215233603,244140623,263671873,284765623,307546873,332150623,358722673,387420487,439453123,474609373,512578123,553584373,597871123,645700813,732421873,791015623,854296873,922640623,996451873,};int main(){ int n; while(~scanf("%d",&n)){ int ans = 0; for(int i=0;i<137;i++) if(a[i]==n) ans = 1; if(ans) puts("Yes"); else puts("No"); } return 0;}