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  1. #include <stdio.h>
  2. #include <math.h>
  3.  
  4. // 判断是否为素数的函数
  5. int is_prime(unsigned long long n) {
  6.     if (n <= 1) return 0;
  7.     if (n == 2) return 1;
  8.     if (n % 2 == 0) return 0;
  9.  
  10.     unsigned long long limit = (unsigned long long)sqrt(n) + 1;
  11.     for (unsigned long long i = 3; i <= limit; i += 2) {
  12.         if (n % i == 0) {
  13.             return 0;
  14.         }
  15.     }
  16.     return 1;
  17. }
  18.  
  19. int main() {
  20.     const unsigned long long limit = 10000000000; // 上限1亿
  21.     unsigned long long perfect_count = 0;
  22.  
  23.     // 已知完美数对应的素数p列表
  24.     int primes[] = {2, 3, 5, 7, 13, 17, 19, 31};
  25.     int num_primes = sizeof(primes) / sizeof(primes[0]);
  26.  
  27.     printf("在0~%llu范围内的完美数有：\n", limit);
  28.  
  29.     for (int i = 0; i < num_primes; i++) {
  30.         int p = primes[i];
  31.         // 计算梅森素数：2^p - 1
  32.         unsigned long long mersenne = (1ULL << p) - 1;
  33.  
  34.         // 计算对应的完美数：(2^(p-1)) * (2^p - 1)
  35.         unsigned long long perfect_number = (1ULL << (p - 1)) * mersenne;
  36.  
  37.         // 检查完美数是否超过上限
  38.         if (perfect_number > limit) {
  39.             break;
  40.         }
  41.  
  42.         // 检查梅森数是否为素数
  43.         if (is_prime(mersenne)) {
  44.             printf("%llu\n", perfect_number);
  45.             perfect_count++;
  46.         }
  47.     }
  48.  
  49.     printf("共找到%llu个完美数\n", perfect_count);
  50.     return 0;
  51. }
Success #stdin #stdout 0s 5320KB
comments (?)
stdin 
Standard input is empty
stdout 
在0~10000000000范围内的完美数有：
6
28
496
8128
33550336
8589869056
共找到6个完美数
https://ideone.com/R1iAp1
language:
C (gcc 8.3)
created:
2 days ago
visibility:
public

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