标签：rms 题目 class -o clu asc ogr += int
The first line of the input is T(1≤ T ≤ 100), which stands for the number of test cases you need to solve.
Each test case contains a line with a positive integer N (2 ≤ N ≤ 10^5).
For each test case, print the case number and determine whether or not the number is perfect.
If the number is perfect, display the sum of its positive divisors less than itself. The ordering of the
terms of the sum must be in ascending order. If a number is not perfect, print "Not perfect.".
3 6 8 28
Case 1: 6 = 1 + 2 + 3 Case 2: Not perfect. Case 3: 28 = 1 + 2 + 4 + 7 + 14
解题思路：测试数据不大，暴力水过！
AC代码一：
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn=1e5+5; 4 int t,n,sum,k,s[maxn]; 5 int main(){ 6 while(~scanf("%d",&t)){ 7 for(int i=1;i<=t;++i){ 8 sum=k=0; 9 scanf("%d",&n); 10 for(int j=1;j<=n/2;++j) 11 if(n%j==0){sum+=j;s[k++]=j;} 12 printf("Case %d: ",i); 13 if(sum!=n)printf("Not perfect.\n"); 14 else{ 15 printf("%d = %d",n,s[0]); 16 for(int j=1;j<k;++j) 17 printf(" + %d",s[j]); 18 printf("\n"); 19 } 20 } 21 } 22 return 0; 23 }
AC代码二：也可以先打表找出10^5内所有完数，一共就4个完数：6，28，496，8128，同样水过！
1 #include<cstdio> 2 int t,n; 3 int main(){ 4 while(~scanf("%d",&t)){ 5 for(int i=1;i<=t;++i){ 6 scanf("%d",&n); 7 printf("Case %d: ",i); 8 if(n==6)printf("6 = 1 + 2 + 3\n"); 9 else if(n==28)printf("28 = 1 + 2 + 4 + 7 + 14\n"); 10 else if(n==496)printf("496 = 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248\n"); 11 else if(n==8128)printf("8128 = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 127 + 254 + 508 + 1016 + 2032 + 4064\n"); 12 else printf("Not perfect.\n"); 13 } 14 } 15 return 0; 16 }
2018 ACM 国际大学生程序设计竞赛上海大都会赛重现赛-B-Perfect Numbers（完数）
标签：rms 题目 class -o clu asc ogr += int
原文地址：https://www.cnblogs.com/acgoto/p/9452519.html
