hdu 4565

Problem Description

　　A sequence S_n is defined as:

Where a, b, n, m are positive integers.┌x┐is the ceil of x. For example, ┌3.14┐=4. You are to calculate S_n.
　　You, a top coder, say: So easy! 

 

Input

　　There are several test cases, each test case in one line contains four positive integers: a, b, n, m. Where 0< a, m < 2¹⁵, (a-1)²< b < a², 0 < b, n < 2³¹.The input will finish with the end of file.

 

Output

　　For each the case, output an integer S_n.

 

Sample Input

2 3 1 2013
2 3 2 2013
2 2 1 2013

 

Sample Output

4
14
4

 

 (a+sqrt(b)）^n向上取整%M
 令A(n)=(a+sqrt(b))^n,B(n)=(a-sqrt(b))^n
 易得C(n)=A(n)+B(n)为整数
 例如：2.3+0.7 ，由于（a-)^<b<a^,因此B(n)为小于1的小数
 那么A(n)向上取整的结果就是C(n),题目也就是求C(n)%M

 #include <iostream>
 #include <cstring>
 #include <string>
 #include <queue>
 #include <set>
 #include <cmath>
 #include <cstdio>
 #include <algorithm>
 #include <cstdlib>
 #define ll long long
 #define max(x,y) (x>y?x:y)
 #define min(x,y)  (x<y?x:y)
 #define gep(i,a,b) for(ll i=a;i<=b;i++)
 using namespace std;
 ll a,b,n,mod;
/*
矩阵乘法的相乘矩阵的行数、列数都要一样，这和行列式不同。
*/
 struct ma{
     ll m[][];
     ma(){
         memset(m,,sizeof(m));
     }
 };
 ma qu(ma a,ma b,ll mod){
     ma c;
     gep(k,,){
         gep(i,,){
                 if(a.m[i][k]){
                     gep(j,,){
                     c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j]%mod+mod)%mod;
                     }
             }
         }
     }
     return c;
 }
 ma quick_qu(ma a,ll b,ll mod){
     ma c;
     gep(i,,){
         c.m[i][i]=1ll;
     }
     while(b){
         if(b&) c=qu(c,a,mod);
         b>>=;
         a=qu(a,a,mod);
     }
     return c;
 }
 int main()
 {
     while(~scanf("%lld%lld%lld%lld",&a,&b,&n,&mod)){
         if(n==){
             printf("%lld\n",*a%mod);
             continue;
         }
          ma c;
          c.m[][]=*a;c.m[][]=b-a*a;
          c.m[][]=;c.m[][]=;
          ma d;
          d.m[][]=*a*a*a+*a*b;d.m[][]=*a*a+*b;d.m[][]=*a;
          ma e=quick_qu(c,n-,mod);
          ma f;
          f=qu(e,d,mod);     //矩阵乘法不满足交换律，e,d位置不能换。
          printf("%lld\n",f.m[][]);
     }
     return ;
 }