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

09-09 23:14
查看更多