输入正整数n(2<=n<=100),把阶乘n!=1*2*3*4*...*n分解成素因子相乘的形式,从小到大输出各个素数(2, 3, 5, ...)的指数。例如825=3*5^2*11应表示成(0, 1, 2, 0, 1),表示分别有0,1,2,0,1个2, 3, 5, 7, 11。你的程序应忽略比最大素因子更大的素数(否则末尾会有无穷多个0)。
如5! = 3, 1, 1
53! = 49, 23, 12, 8, 4, 4, 3, 2, 2, 1, 1, 1, 1, 1, 1, 1
#include <iostream> #include <string.h> using namespace std; int prime[100]; int count = 0; bool isPrime(int a) { for (int i = 2; i*i <= a; i++) { if (a%i == 0) { return false; } } return true; } void constructPrimeTable(int a) { for (int i = 2; i <= 100; i++) { if (isPrime(i)) { prime[count++] = i; } } } void fun(int n) { constructPrimeTable(n); int p[100]; memset(p, 0, sizeof(int)*100); int maxP = 0; for (int i = 2; i <= n; i++) { int temp = i; for (int j = 0; j < count; j++) { if (temp == 1) { break; } while ((temp%prime[j]) == 0) { temp /= prime[j]; p[j]++; if (j > maxP) { maxP = j; } } } } for (int i = 0; i <= maxP; i++) { cout << p[i] << ", "; } } int main() { fun(5); cout << endl; fun(53); cout << endl; return 0; }