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?
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
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; }