玄学,少开一个数组就TLE,多开一个数组就MLE
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玄学,少开一个数组就TLE,多开一个数组就MLE
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woshiren楼主2020/5/8 10:48

这常数卡得也太过头了吧?

#include<cstdio>
#define int long long
using namespace std;
int mul[5000005],ans,n,p,k,pre[5000005],last[5000005];
void exgcd(int a,int b,int &x,int &y)
{
    if (b==0)
    {
        x=1,y=0;
        return ;   
    }
    exgcd(b,a%b,y,x);
    y-=(a/b)*x;
}
inline int read()
{
	int X=0; bool flag=1; char ch=getchar();
	while(ch<'0'||ch>'9') {if(ch=='-') flag=0; ch=getchar();}
	while(ch>='0'&&ch<='9') {X=(X<<1)+(X<<3)+ch-'0'; ch=getchar();}
	if(flag) return X;
	return ~(X-1);
}
signed main()
{
    n=read();p=read();k=read();
    mul[0]=1;
    for (int i=1;i<=n;i++)
        mul[i]=mul[i-1]*k%p;
    int s=1,x,y;pre[0]=1;last[n+1]=1;
    for (int i=1;i<=n;i++)
    {
        last[i]=read();
        s*=last[i];s%=p;
        pre[i]=pre[i-1]*last[i]%p;
    }
    for (int i=n;i>=1;i--)
        last[i]=last[i+1]*last[i]%p;
    exgcd(s,p,x,y);
    x=(x%p+p)%p;
    for (int i=1;i<=n;i++)
        ans=(ans%p+mul[i]%p*pre[i-1]%p*last[i+1]%p)%p;
    printf("%lld",ans*(x%p+p)%p);
    return 0;
}
2020/5/8 10:48
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