比赛 20251022赛前模拟1 评测结果 AAAAAAAAAA
题目名称 学姐的巧克力盒 最终得分 100
用户昵称 李金泽 运行时间 6.317 s
代码语言 C++ 内存使用 14.30 MiB
提交时间 2025-10-22 11:39:54
显示代码纯文本
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#define N 1000005
#define ls(x) x<<1
#define rs(x) x<<1|1
#define fo(i,l,r) for(int i=l;i<=r;i++)
#define rf(i,r,l) for(int i=r;i>=l;i--)
#define ll long long
using namespace std;
int n,m,op,x,y;ll k,p1,p2,p,a[N],t1[N<<2],t2[N<<2];
ll phi(ll x)
{
    ll ans=x;
    for(ll i=2;i*i<=x;i++)                           
        if(!(x%i))
        {
            ans*=i-1;ans/=i;
            while(!(x%i))x/=i;
        }
    if(x>1)ans*=x-1,ans/=x;
    return ans;
}
void pu(int x)
{
    if(p1)t1[x]=t1[ls(x)]*t1[rs(x)]%p1;
    if(p2)t2[x]=t2[ls(x)]*t2[rs(x)]%p;
}
void bd(int l,int r,int x)
{
    if(l==r){if(p1)t1[x]=a[l]%p1;if(p2)t2[x]=a[l]%p;return;}
    int mid=l+r>>1;
    bd(l,mid,ls(x));bd(mid+1,r,rs(x));
    pu(x);
}
ll q1(int l,int r,int sl,int sr,int x)
{
    if(sl<=l&&r<=sr)return t1[x];
    int mid=l+r>>1;ll sum=1;
    if(sl<=mid)sum=q1(l,mid,sl,sr,ls(x));
    if(mid<sr)sum=sum*q1(mid+1,r,sl,sr,rs(x))%p1;
    return sum;
}
ll q2(int l,int r,int sl,int sr,int x)
{
    if(sl<=l&&r<=sr)return t2[x];
    int mid=l+r>>1;ll sum=1;
    if(sl<=mid)sum=q2(l,mid,sl,sr,ls(x));
    if(mid<sr)sum=sum*q2(mid+1,r,sl,sr,rs(x))%p;
    return sum;
}
ll fp(ll a,ll n)
{
    ll ans=1;a%=p2;
    while(n)
    {
        if(n&1)ans=ans*a%p2;
        a=a*a%p2;
        n>>=1;
    }
    return ans;
}
void swap(int &x,int &y){int t=x;x=y;y=t;}
int max(int x,int y){return x>y?x:y;}
int min(int x,int y){return x<y?x:y;}
int read(){
	int sum=0;bool f=0;char c=getchar();
	for(;c<48||c>57;c=getchar())if(c==45)f=1;
	for(;c>=48&&c<=57;c=getchar())sum=sum*10+(c&15);
	return f?-sum:sum;
}
int main(){
	freopen("chocolatebox.in","r",stdin);freopen("chocolatebox.out","w",stdout);
	n=read();m=read();k=read();p1=read();p2=read();if(p2)p=phi(p2);
	fo(i,1,n)a[i]=read();
	bd(1,n,1);
	while(m--)
	{
	    op=read();x=read();y=read();
	    if(op==1)printf("%lld\n",q1(1,n,x,y,1));
	    else printf("%lld\n",fp(k,q2(1,n,x,y,1)));
    }
	return 0;
}