我正在LCM上解决以下问题:计算N个数字的LCM取模1000000007
我的方法:
typedef unsigned long long ull;
const ull mod=1000000007;
ull A[10009];
/*Euclidean GCD*/
ull gcd(ull a,ull b)
{
while( b != 0)
{
ull t = b;
b= a %t;
a = t;
}
return a;
}
ull lcm(ull a, ull b)
{
return (a/gcd(a,b))%mod*(b%mod);
}
ull lcms(int l ,ull * A)
{
int i;
ull result;
result = 1;
for (i = 0; i < l; i++)
result = lcm(result, A[i])%1000000007;
return result;
}
int main()
{
int T;
cin>>T;
while(T--)/*Number of test cases*/
{
int N;
cin>>N;/*How many Numbers in Array*/
for(int i=0;i<N;++i)
{
cin>>A[i];//Input Array
}
cout<<lcms(N,A)%1000000007<<endl;
}
return 0;
}
提交解决方案时,我得到了错误答案。
约束是:
1<=N<=1000
and 1<=A[i]<=10000
AT IDEONE
我猜我因为溢出而得到错误答案。如何改善我的守则?
谢谢!
最佳答案
1000000007
太大了,我无法举个例子。我以17
为例:
LCMS(10, 9, 8) % 17 =
LCM(10, LCM(9, 8)) % 17 =
LCM(10, 72) % 17 =
360 % 17 =
3
这是您的代码执行的操作:
LCMS(10, 9, 8) % 17 =
LCM(10, LCM(9, 8) % 17) % 17 =
LCM(10, 72 % 17) % 17 =
LCM(10, 4) % 17 =
40 % 17 =
6
错了
ALSO AT IDEONE