题目描述
We consider a positive integer perfect, if and only if it is equal to the sum of its positive divisors less than itself.
For example, 6 is perfect because 6 = 1 + 2 + 3.
Could you write a program to determine if a given number is perfect or not?
输入描述:
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).
输出描述:
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 theterms of the sum must be in ascending order. If a number is not perfect, print "Not perfect.".
示例1
输入
3 6 8 28
输出
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 }
原文地址：https://www.cnblogs.com/acgoto/p/9452519.html
时间： 2024-08-02 22:33:02