• D


    题意:给你一个数组a,q次查询,每次l,r,要求 (a_{l}^{a_{l+1}}^{a_{l+2}}...{a_r})
    题解:由欧拉降幂可知,最多log次eu(m)肯定变1,那么直接暴力即可,还有一个问题是欧拉降幂公式,
    (a^{b}mod c=a^{b modphi(c)+phi(c)}mod c(a>c))
    需要改写快速幂

    //#pragma GCC optimize(2)
    //#pragma GCC optimize(3)
    //#pragma GCC optimize(4)
    //#pragma GCC optimize("unroll-loops")
    //#pragma comment(linker, "/stack:200000000")
    //#pragma GCC optimize("Ofast,no-stack-protector")
    //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
    #include<bits/stdc++.h>
    #define fi first
    #define se second
    #define db double
    #define mp make_pair
    #define pb push_back
    #define pi acos(-1.0)
    #define ll long long
    #define vi vector<int>
    #define mod 1000000007
    #define ld long double
    //#define C 0.5772156649
    //#define ls l,m,rt<<1
    //#define rs m+1,r,rt<<1|1
    #define pll pair<ll,ll>
    #define pil pair<int,ll>
    #define pli pair<ll,int>
    #define pii pair<int,int>
    #define ull unsigned long long
    //#define base 1000000000000000000
    #define fin freopen("a.txt","r",stdin)
    #define fout freopen("a.txt","w",stdout)
    #define fio ios::sync_with_stdio(false);cin.tie(0)
    inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
    inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
    inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
    template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
    template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
    inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
    inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=(ans*a>=c?ans*a%c+c:ans*a);a=(a*a>=c?a*a%c+c:a*a),b>>=1;}return ans;}
    
    using namespace std;
    
    const ull ba=233;
    const db eps=1e-5;
    const ll INF=0x3f3f3f3f3f3f3f3f;
    const int N=100000+10,maxn=100000+10,inf=0x3f3f3f3f;
    
    ll eu(ll n)
    {
        ll ans=n;
        for(ll i=2;i*i<=n;i++)
        {
            if(n%i==0)
            {
                ans=ans/i*(i-1);
                while(n%i==0)n/=i;
            }
        }
        if(n!=1)ans=ans/n*(n-1);
        return ans;
    }
    ll a[N],b[40];
    int main()
    {
        int n;ll m;
        scanf("%d%lld",&n,&m);
        b[0]=m;
        for(int i=1;i<=35;i++)b[i]=eu(b[i-1]);
        for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
        int q;scanf("%d",&q);
        while(q--)
        {
            int l,r;scanf("%d%d",&l,&r);
            if(r-l+1>30)r=l+30;
            ll ans=1;
            for(int i=r,j=r-l;i>=l;i--,j--)
            {
                ans=qp(a[i],ans,b[j]);
    //            printf("%d %lld %lld
    ",i,b[j],ans);
            }
            printf("%lld
    ",ans%b[0]);
        }
        return 0;
    }
    /********************
    2 18
    2 2
    1
    1 2
    ********************/
    
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  • 原文地址:https://www.cnblogs.com/acjiumeng/p/10622156.html
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