uva 10791 - Minimum Sum LCM(分解因子）

题目大意：uva 10791 - Minimum Sum LCM

题目大意：给出一个n，将n分解成n = p1 ^ k1 * p2 ^ k2 * ... * pm ^ km，然后求解sum = ∑(1≤i≤m)pi ^ ki.

解题思路：比较费解的是n本身就是素数，那么n应该分解成n^ 1 + 1 ^ 1,所以sum= n + 1，还有一种就是n = p ^ k，sum = p ^ k + 1.

#include <stdio.h>#include <math.h>int main () {int cas = 1, n;while (scanf("%d", &n), n) {long long sum = 0, tmp = sqrt(n), cnt = 0;for (int i = 2; i <= tmp; i++) {if (n % i == 0) {int c = 1;cnt++;while (n % i == 0) {c *= i;n /= i;}sum += c;}}if (n > 1 || cnt == 0) {sum += n;cnt++;}if (cnt == 1)sum++;printf("Case %d: %lld\n", cas++, sum);}return 0;}