hdu - 2028 - Lowest Common Multiple Plus
2013-02-08 06:44
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题意:求n个数的最小公倍数。
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2028
——>>两两归并就好。
#include <cstdio>
using namespace std;
const int maxn = 1000000 + 10;
long long a[maxn];
long long gcd(long long x, long long y)
{
return (x % y == 0) ? y : gcd(y, x % y);
}
int main()
{
int n, i;
long long lcd;
while(~scanf("%d", &n))
{
for(i = 0; i < n; i++)
scanf("%I64d", &a[i]);
if(n == 1) printf("%I64d\n", a[0]);
else
{
lcd = a[1] / gcd(a[1], a[0]) * a[0];
for(i = 2; i < n; i++) lcd = a[i] / gcd(a[i], lcd) * lcd;
printf("%I64d\n", lcd);
}
}
return 0;
}
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2028
——>>两两归并就好。
#include <cstdio>
using namespace std;
const int maxn = 1000000 + 10;
long long a[maxn];
long long gcd(long long x, long long y)
{
return (x % y == 0) ? y : gcd(y, x % y);
}
int main()
{
int n, i;
long long lcd;
while(~scanf("%d", &n))
{
for(i = 0; i < n; i++)
scanf("%I64d", &a[i]);
if(n == 1) printf("%I64d\n", a[0]);
else
{
lcd = a[1] / gcd(a[1], a[0]) * a[0];
for(i = 2; i < n; i++) lcd = a[i] / gcd(a[i], lcd) * lcd;
printf("%I64d\n", lcd);
}
}
return 0;
}
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