九度OJ 1056--最大公约数 1439--Least Common Multiple 【辗转相除法】
题目地址:http://ac.jobdu.com/problem.php?pid=1056
- 题目描述:
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输入两个正整数,求其最大公约数。
- 输入:
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测试数据有多组,每组输入两个正整数。
- 输出:
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对于每组输入,请输出其最大公约数。
- 样例输入:
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49 14
- 样例输出:
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7
#include <stdio.h> int gcd1 (int a, int b){ if (b == 0) return a; else return gcd1 (b, a % b); } int gcd2 (int a, int b){ int tmp; while (b != 0){ tmp = a; a = b; b = tmp % b; } return a; } int main (void){ int a, b; while (scanf ("%d%d", &a, &b) != EOF){ printf ("%d\n", gcd1 (a, b)); printf ("%d\n", gcd2 (a, b)); } return 0; }
题目地址:http://ac.jobdu.com/problem.php?pid=1439
- 题目描述:
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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.
- 输入:
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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.
- 输出:
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For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.
- 样例输入:
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2 3 5 7 15 6 4 10296 936 1287 792 1
- 样例输出:
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105 10296
#include <stdio.h> int Gcd (int a, int b){ int tmp; while (b != 0){ tmp = a; a = b; b = tmp % b; } return a; } int main(void){ int n; int a; int b; int m; while (scanf ("%d", &n) != EOF){ while (n-- != 0){ scanf ("%d%d", &m, &a); --m; while (m != 0){ scanf ("%d", &b); a = a / Gcd (a, b) * b; --m; } printf ("%d\n", a); } } return 0; }
参考资料:维基百科 -- 辗转相除法