So Easy!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 813 Accepted Submission(s): 226

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 < 2

¹⁵, (a-1)

²< b < a

², 0 < b, n < 2

³¹.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 

详解在这:

另外Cn=(a+sqrt(b))^n+(a-sqrt(b))^n =>Cn*(a+sqrt(b) + a-sqrt(b))=(a+sqrt(b))^(n+1)+(a-sqrt(b))^(n+1)+(a*a-b)*(a+sqrt(n))^(n-1)+(a*a-b)*(a-sqrt(b))^(n-1)=C(n+1) + (a*a-b)C(n-1) =>C(n+1)=2a*Cn+(b-a^2)*C(n-1)

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             #define INF 99999999using namespace std;const int MAX=10;__int64 array[2][2],sum[2][2];void MatrixMult(__int64 a[2][2],__int64 b[2][2],__int64 mod){ __int64 c[2][2]={0}; for(int i=0;i<2;++i){ for(int j=0;j<2;++j){ for(int k=0;k<2;++k){ c[i][j]+=a[i][k]*b[k][j]; } } } for(int i=0;i<2;++i){ for(int j=0;j<2;++j)a[i][j]=c[i][j]%mod; }}__int64 Matrix(__int64 a,__int64 b,__int64 k,__int64 mod){ array[0][0]=a%mod,array[0][1]=(b%mod+mod)%mod; array[1][0]=1,array[1][1]=0; sum[0][0]=sum[1][1]=1; sum[0][1]=sum[1][0]=0; while(k){ if(k&1)MatrixMult(sum,array,mod); MatrixMult(array,array,mod); k>>=1; } return (a*sum[0][0]+2*sum[0][1])%mod;}int main(){ __int64 a,b,n,m; while(scanf("%I64d%I64d%I64d%I64d",&a,&b,&n,&m)!=EOF){ if(n == 0)printf("I64d\n",1%m); else if(n == 1)printf("%I64d\n",2*a%m); else printf("%I64d\n",Matrix(2*a,b-a*a,n-1,m)); } return 0;}