分块计算

题目：http://acm.upc.edu.cn/problem.php?id=2219

 

 

题意：

It's easy for ACMer to calculate A^X mod P. Now given seven integers n, A, K, a, b, m, P, and a

function f(x) which defined as following.

f(x) = K, x = 1

f(x) = (a*f(x-1) + b)%m , x > 1

Now, Your task is to calculate

( A^(f(1)) + A^(f(2)) + A^(f(3)) + ...... + A^(f(n)) ) modular P. 

 

本题有点技巧，不要直接快速幂运算，这里要分两部分，详见代码：

#include <iostream>
#include <string.h>
#include <stdio.h>

using namespace std;
typedef long long LL;

LL f[1000005];
LL x1[1000005];
LL x2[1000005];

LL quick_mod(LL a,LL b,LL m)
{
    LL ans=1;
    while(b)
    {
        if(b&1)
        {
            ans=ans*a%m;
            b--;
        }
        b>>=1;
        a=a*a%m;
    }
    return ans;
}

LL multi(LL n,LL p)
{
    return x1[n/50000]*x2[n%50000]%p;
}

int main()
{
    LL t,n,A,K,a,b,m,p,tt=1;
    LL ans;
    cin>>t;
    while(t--)
    {
        cin>>n>>A>>K>>a>>b>>m>>p;
        A%=p;a%=m;b%=m;
        f[1]=K;
        for(int i=2;i<=n;i++)
            f[i]=(f[i-1]%m*a%m+(b%m))%m;
        LL val=quick_mod(A,50000,p);
        x1[0]=1;
        for(int i=1;i<50001;i++)
            x1[i]=(val%p)*x1[i-1]%p;
        x2[0]=1;
        for(int i=1;i<50001;i++)
            x2[i]=(A%p)*x2[i-1]%p;
        ans=0;
        for(int i=1;i<=n;i++)
        {
            ans+=multi(f[i],p);
            ans%=p;
        }
        cout<<"Case #"<<tt++<<": ";
        cout<<ans<<endl;
    }
    return 0;
}