杭电acm 1019 Least Common Multiple
2012-10-05 18:26
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本题是求所给数的最小公倍数,方法:先求gcm,再求lcm,基本方法和平常相同,不同的是处理数据。第一次输入要存储起来作为前者,以后的每次输入都要和“前者”求一次lcm,作为下一次的前者,一直这样,最终求出lcm
#include<iostream>
using namespace std;
int getg(int a,int b);//get gcm
int getl(int a,int b);//get lcm
int main()
{
int t,n,m,s;
cin>>t;
while(t--)
{
cin>>n;
for(int i=0;i<n;++i)
{
cin>>m;
if(i==0)
s=m;//第一次先用一个变量储存下来
else
s=getl(s,m);
}
cout<<s<<endl;
}
return 0;
}
int getg(int a,int b)
{
int r;
if(a<b)
{
r=a;
a=b;
b=r;
}
while(b)
{
r=a%b;
a=b;
b=r;
}
return a;
}
int getl(int a,int b)
{
return a/getg(a,b)*b;//这个相当重要,不用a*b/getg(a,b)防止超出int型
}
#include<iostream>
using namespace std;
int getg(int a,int b);//get gcm
int getl(int a,int b);//get lcm
int main()
{
int t,n,m,s;
cin>>t;
while(t--)
{
cin>>n;
for(int i=0;i<n;++i)
{
cin>>m;
if(i==0)
s=m;//第一次先用一个变量储存下来
else
s=getl(s,m);
}
cout<<s<<endl;
}
return 0;
}
int getg(int a,int b)
{
int r;
if(a<b)
{
r=a;
a=b;
b=r;
}
while(b)
{
r=a%b;
a=b;
b=r;
}
return a;
}
int getl(int a,int b)
{
return a/getg(a,b)*b;//这个相当重要,不用a*b/getg(a,b)防止超出int型
}
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