hdu 1019 Least Common Multiple
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题意要求n个数的lcm,而对于任意的两个数a,b来说 LCM(a,b)=a*b/gcd(a,b),其中gcd(a,b)是a,b的最大公约数,我们利用这个公式就可以很容易的得出答案
PS:一开始我没有处理m=1的情况导致程序超时,这是我自己编程不严谨造成的,下次一定改正。
#include <cstdio>#include <cmath>#include <iostream>using namespace std;long long gcd(long long a,long long b){ if (b==0) return a; else return gcd(b,a%b);}int main(){ int N; while (scanf("%d",&N)!=EOF) { while (N--) { int m; scanf("%d",&m); long long a,b; a=1; while (m--) { scanf("%I64d",&b); a=a/gcd(a,b)*b; } printf("%I64d\n",a); } } return 0;}
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