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1019
Least Common Multiple Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 28696 Accepted Submission(s): 10785 Problem Description 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. Input 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. Output For 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
#include<iostream> #include<algorithm> #include<cmath> using namespace std; int cmp(int a,int b) { return a<b; } int gcd(int a,int b) { if(!b) return a; return gcd(b,a%b); } int LCM(int a,int b) { return (a*b)/gcd(a,b); } int main() { int n,i,m,mm; int a[100000]; cin>>n; while(n--) { cin>>m; for(i=0;i<m;i++) cin>>a[i]; sort(a,a+m,cmp); if(m==1) cout<<a[0]; if(m>1) { mm=a[0]; for(i=1;i<m;i++) { mm=LCM(mm,a[i]); } cout<<mm<<endl; } } return 0; }
上面是我的代码 WA
下面还是我的代码 AC
#include<iostream> #include<algorithm> #include<cmath> using namespace std; int cmp(int a,int b) { return a<b; } int gcd(int a,int b) { if(!b) return a; return gcd(b,a%b); } int LCM(int a,int b) { return a/gcd(a,b)*b; } int main() { int n,i,m,mm; int a[100000]; cin>>n; while(n--) { cin>>m; for(i=0;i<m;i++) cin>>a[i]; sort(a,a+m,cmp); if(m==1) cout<<a[0]<<endl; if(m>1) { mm=a[0]; for(i=1;i<m;i++) { mm=LCM(mm,a[i]); } cout<<mm<<endl; } } return 0; }
与上面代码不同的是,下面的代码在求最小公倍数的时候不是用的(a*b)/gcd(a,b)而是用的a/gcd(a,b)*b还有就是改正了一个格式错误:if(m==1)应该是cout<<a[0]<<endl;
第一次的代码少了endl
求gcd的代码比较巧妙!!
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