【九度OJ】題目1439：Least Common Multiple 解題報告

2019-11-06 06:15:53

字體：大中小

來源：轉載

供稿：網友

【九度OJ】題目1439：Least Common Multiple 解題報告

標簽（空格分隔）：九度OJ

原題地址：http://ac.jobdu.com/PRoblem.php?pid=1439

題目描述：

The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7 and 15 is 105.

輸入：

Input will consist of multiple problem instances. The first line of the input will contain a single integer indicating the number of problem instances. Each instance will consist of a single line of the form m n1 n2 n3 … nm where m is the number of integers in the set and n1 … nm are the integers. All integers will be positive and lie within the range of a 32-bit integer.

輸出：

For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.

樣例輸入：

23 5 7 156 4 10296 936 1287 792 1

樣例輸出：

10510296

Ways

BigInteger類好！

這個題的意思很簡單，其實就是求指定數字的最小公倍數。

我們可以利用上一題的經驗，求出m個數的共同最小公倍數即可。

做題時一個錯誤的地方就是注意兩層循環的嵌套，把循環變量給寫錯了，導致一直出錯。

import java.util.*;import java.math.*;public class Main{ public static void main(String[] args) { Scanner scanner = new Scanner(System.in); String n = scanner.nextLine(); for (int i = 0; i < Integer.parseInt(n); i++) { String line = scanner.nextLine(); String[] params = line.split(" "); BigInteger a = new BigInteger(params[1]); for (int j = 2; j < params.length; j++) { BigInteger b = new BigInteger(params[j]);//是j，不是i a = a.multiply(b).divide(a.gcd(b)); } System.out.println(a.toString()); } }}

本來以為C++的版本也會同樣的容易，可是還是遇到點問題，很不爽。原因是因為結果超出了int范圍。改成long long就好了。

#include <stdio.h>long long gcd(long long a, long long b) { return b != 0 ? gcd(b, a % b) : a;}int main() { int n; while (scanf("%d", &n) != EOF) { while (n-- != 0) { int m; scanf("%d", &m); long long answer = 1; while (m-- != 0) { int temp; scanf("%d", &temp); answer = answer * temp / gcd(answer, temp); } printf("%lld/n", answer); } } return 0;}