1 条题解
-
0
C :
#include <stdio.h> #include <stdlib.h> #define N 1000 #define P 10000 int prime(int*); // 求质数表 int factor(int*, int, int*); // 求factor (因子) int fsum(int*, int); // sum ot proper factor int main() { int ptable[N+1] = {0}; // 储存质数表 int fact[N+1] = {0}; // 储存因式分解结果 int count1, count2, i; count1 = prime(ptable); for(i = 0; i <= P; i++) { count2 = factor(ptable, i, fact); if(i == fsum(fact, count2)) printf("%d\n", i); } printf("\n"); return 0; } int prime(int* pNum) { int i, j; int prime[N+1]; for(i = 2; i <= N; i++) prime[i] = 1; for(i = 2; i*i <= N; i++) { if(prime[i] == 1) { for(j = 2*i; j <= N; j++) { if(j % i == 0) prime[j] = 0; } } } for(i = 2, j = 0; i < N; i++) { if(prime[i] == 1) pNum[j++] = i; } return j; } int factor(int* table, int num, int* frecord) { int i, k; for(i = 0, k = 0; table[i] * table[i] <= num;) { if(num % table[i] == 0) { frecord[k] = table[i]; k++; num /= table[i]; } else i++; } frecord[k] = num; return k+1; } int fsum(int* farr, int c) { int i, r, s, q; i = 0; r = 1; s = 1; q = 1; while(i < c) { do{ r *= farr[i]; q += r; i++; } while(i < c-1 && farr[i-1] == farr[i]); s *= q; r = 1; q = 1; } return s / 2; }
- 1
信息
- ID
- 1357
- 时间
- 1000ms
- 内存
- 2MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者