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hdu-5895 Mathematician QSC(数学)

2024-08-13 23:12:42   212人阅读

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(n−2)+2∗f(n−1)(n≥2)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(n∗y)%(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：
 
难点在怎么退出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(‘\n‘);}  //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\n",ans);    }    return 0;}

　　

 

hdu-5895 Mathematician QSC(数学)


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