UVA 10791 - Minimum Sum LCM(数论)

最新推荐文章于 2018-06-02 20:00:15 发布

原创最新推荐文章于 2018-06-02 20:00:15 发布 · 1.3k 阅读

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本文介绍了一种求解给定整数n的最小LCM（Least Common Multiple）的方法，即找到至少两个正整数的LCM等于n，并使它们的和最小。详细解释了求解过程，并提供了代码实现。

Minimum Sum LCM

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LCM (Least Common Multiple) of a set of integers is defined as the minimum number, which is a multiple of all integers of that set. It is interesting to note that any positive integer can be expressed as the LCM of a set of positive integers. For example 12 can be expressed as the LCM of 1, 12 or12, 12 or 3, 4 or 4, 6 or 1, 2, 3, 4 etc.

In this problem, you will be given a positive integer N. You have to find out a set of at least two positive integers whose LCM is N. As infinite such sequences are possible, you have to pick the sequence whose summation of elements is minimum. We will be quite happy if you just print the summation of the elements of this set. So, for N = 12, you should print4+3 = 7 as LCM of 4 and 3 is 12 and 7 is the minimum possible summation.

Sample Output

 
Case 1: 7
Case 2: 7
Case 3: 6

题意：给定一个n，求最小LCM。。LCM求法为，把n分解成几个乘数相加，要求这些乘数互质。

思路：先分解成全是质数p1^n1*p2^n2....pn^nn。最小和为p1^n1+p2^n2+...+pn^nn。

注意几个WA点。要用longlong。不然当输入为2147483647会输出2147483648.超范围。还有如果不能分解或者只能分解出一个因数答案应该为n+1.

代码：

#include <stdio.h>
#include <string.h>
#include <math.h>

long long n, ans, i, j, bo;
int main() {
	int t = 0;
	while (~scanf("%d", &n) && n) {
		long long save = n;
		ans = bo = 0;
		long long num = sqrt(n);
		for (i = 2; i <= num; i ++) {
			long long sum = 1;
			if (n % i == 0) {
				bo ++;
				while (n % i == 0) {
					n /= i;
					sum *= i;
				}
				ans += sum;
			}
		}
		if (n != 1)
			ans += n;
		if (bo < 2) {
			ans = save + 1;
		}
		printf("Case %d: %lld\n", ++ t, ans);
	}
	return 0;
}