Minimum Sum LCM(分解因子）
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.
[cpp][/cpp] view plaincopy
- #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;
- }
