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
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?
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
#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;
}