HDU1019 Least Common Multiple(多个数的最小公倍数)
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InputInput 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.
OutputFor each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.
Sample Input
2 3 5 7 15 6 4 10296 936 1287 792 1
Sample Output
105 10296
这道题挺简单的,但我居然想着去分解质因数然后乘一遍,又果断的TLE了.但其实真的很简单,不断地读入一个数,然后求它和之前所有数的lcm,这个怎么求呢?除去两数的gcd即可.说起来还是太菜了....
代码如下:
#include<cstdio> using namespace std; long long ans=1,n,i,x; long long gcd(long long a,long long b) { if(b>a) { long long w=a;a=b;b=w; } if(a%b==0) { return b; } else { return gcd(b,a%b); } } int main() { long long t; scanf("%lld",&t); while(t--) { ans=1; scanf("%lld",&n); for(i=1;i<=n;i++) { scanf("%lld",&x); ans*=x/gcd(ans,x); } printf("%lld\n",ans); } return 0; }
每日刷题身体棒棒!
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