• hdu-5895 Mathematician QSC(数学)


    题目链接:

    Mathematician QSC

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 189    Accepted Submission(s): 90


    Problem Description
    QSC dream of becoming a mathematician, he believes that everything in this world has a mathematical law.

    Through unremitting efforts, one day he finally found the QSC sequence, it is a very magical sequence, can be calculated by a series of calculations to predict the results of a course of a semester of a student.

    This sequence is such like that, first of all,f(0)=0,f(1)=1,f(n)=f(n2)+2f(n1)(n2)Then the definition of the QSC sequence is g(n)=ni=0f(i)2. If we know the birthday of the student is n, the year at the beginning of the semester is y, the course number x and the course total score s, then the forecast mark is xg(ny)%(s+1).
    QSC sequence published caused a sensation, after a number of students to find out the results of the prediction is very accurate, the shortcoming is the complex calculation. As clever as you are, can you write a program to predict the mark?
     
    Input
    First line is an integer T(1≤T≤1000).

    The next T lines were given n, y, x, s, respectively.

    n、x is 8 bits decimal integer, for example, 00001234.

    y is 4 bits decimal integer, for example, 1234.
    n、x、y are not negetive.

    1≤s≤100000000
     
    Output
    For each test case the output is only one integer number ans in a line.
     
    Sample Input
    2
    20160830 2016 12345678 666
    20101010 2014 03030303 333
     
    Sample Output
    1
    317
     
    题意:
     
    求上面那个式子的值;
     
    思路:
     
    难点在怎么推出g[n]的表达式了;g(n)=f(n)*f(n+1)/2;
     
    f(n)=f(n-2)+2*f(n-1)
    f(n)*f(n-1)=f(n-2)*f(n-1)+2*f(n-1)*f(n-1);
    2*f(n-1)*f(n-1)=f(n)*f(n-1)-f(n-2)*f(n-1);
    连加得到g(n)=f(n)*f(n+1)/2;
    然后就是矩阵快速幂和指数循环节的套路了;
     
    AC代码:
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #include <bits/stdc++.h>
    #include <stack>
    #include <map>
      
    using namespace std;
      
    #define For(i,j,n) for(int i=j;i<=n;i++)
    #define mst(ss,b) memset(ss,b,sizeof(ss));
      
    typedef  long long LL;
      
    template<class T> void read(T&num) {
        char CH; bool F=false;
        for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
        for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
        F && (num=-num);
    }
    int stk[70], tp;
    template<class T> inline void print(T p) {
        if(!p) { puts("0"); return; }
        while(p) stk[++ tp] = p%10, p/=10;
        while(tp) putchar(stk[tp--] + '0');
        putchar('
    ');
    }
      
    //const LL mod=1e9+7;
    const double PI=acos(-1.0);
    const LL inf=1e18;
    const int N=(1<<20)+10;
    const int maxn=1e5+10;
    const double eps=1e-12;
     
    LL prime[maxn],mod;
    int vis[maxn],cnt=0;
    struct matrix 
    {
        LL a[2][2];
    };
    matrix cal(matrix A,matrix B)
    {
        matrix C;
        for(int i=0;i<2;i++)
        {
            for(int j=0;j<=2;j++)
            {
                C.a[i][j]=0;
                for(int k=0;k<2;k++)
                {
                    C.a[i][j]+=A.a[i][k]*B.a[k][j];
                    C.a[i][j]%=mod;
                }
            }
        }
        return C;
    }
    
    LL pow_mod(LL y)
    {
        if(y==0)return 0;
        else if(y==1)return 1;
        else if(y==2)return 2;
        else y-=2;
        matrix s,base;
        s.a[0][0]=s.a[1][1]=1;s.a[0][1]=s.a[1][0]=0;
        base.a[0][0]=2,base.a[0][1]=base.a[1][0]=1,base.a[1][1]=0;
        while(y)
        {
            if(y&1)s=cal(s,base);
            base=cal(base,base);
            y>>=1;
        }
        return (s.a[0][0]*2+s.a[0][1])%mod;
    }
    inline void Init()
    {
        for(int i=2;i<maxn;i++)
        {
            if(!vis[i])
            {
                for(int j=2*i;j<maxn;j+=i)vis[j]=1;
                prime[++cnt]=(LL)i;
            }
        }
    }
    LL phi(LL fx)
    {
        LL s=fx;
        for(int i=1;i<=cnt;i++)
        {
            if(fx<prime[i])break;
            if(fx%prime[i]==0)
            {
                s=s/prime[i]*(prime[i]-1);
                while(fx%prime[i]==0)fx/=prime[i];
            }
        }
        if(fx>1)s=s/fx*(fx-1);
        return s;
    }
    LL powmod(LL a,LL b,LL mo)
    {
      LL s=1,base=a;
      while(b)
      {
        if(b&1)s=s*base%mo;
        base=base*base%mo;
        b>>=1;
      }
      return s;
    }
    int main()
    {
        Init();
        int t;
        LL n,y,x,s;
        read(t);
        while(t--)
        {
            scanf("%lld%lld%lld%lld",&n,&y,&x,&s);
            s++;
            mod=phi(s)*2;
            LL ans=pow_mod(n*y)*pow_mod(n*y+1)%mod/2+mod/2;
            ans=powmod(x,ans,s);
            printf("%lld
    ",ans);
        }
        return 0;
    }
    

      

     
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  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5886051.html
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