So Easy!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2286 Accepted Submission(s): 710
Problem Description
A sequence Sn 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 Sn.
You, a top coder, say: So easy!


Where a, b, n, m are positive integers.┌x┐is the ceil of x. For example, ┌3.14┐=4. You are to calculate Sn.
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 Sn.
Sample Input
2 3 1 2013
2 3 2 2013
2 2 1 2013
Sample Output
4
14
4
挑战程序设计竞赛P268,关键是求初始矩阵:
an+1 = a*an+b*bn;
bn+1 = an+a*bn;
a[0][0],a[0][1]分别是公式1里面的系数,a和b;
a[1][0],a[1][1]分别是公式2里面的系数,1和a;
然后就是矩阵快速幂了:
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<vector> 5 #include<algorithm> 6 #include<cmath> 7 #define M(a,b) memset(a,b,sizeof(a)) 8 typedef long long LL; 9 10 using namespace std; 11 12 13 long long a,b,n,m; 14 15 struct matrix 16 { 17 LL mat[2][2]; 18 void init() 19 { 20 mat[0][0] = a; 21 mat[0][1] = b; 22 mat[1][0] = 1; 23 mat[1][1] = a; 24 } 25 }; 26 27 matrix mamul(matrix aa,matrix bb) 28 { 29 matrix c; 30 for(int i = 0;i<2;i++) 31 { 32 for(int j = 0;j<2;j++) 33 { 34 c.mat[i][j] = 0; 35 for(int k = 0;k<2;k++) 36 c.mat[i][j]+=(aa.mat[i][k]*bb.mat[k][j]); 37 c.mat[i][j]%=m; 38 } 39 } 40 return c; 41 } 42 43 matrix mul(matrix s, int k) 44 { 45 matrix ans; 46 ans.init(); 47 while(k>=1) 48 { 49 if(k&1) 50 ans = mamul(ans,s); 51 k = k>>1; 52 s = mamul(s,s); 53 } 54 return ans; 55 } 56 57 int main() 58 { 59 while(scanf("%I64d%I64d%I64d%I64d",&a,&b,&n,&m)==4) 60 { 61 matrix ans; 62 ans.init(); 63 ans = mul(ans,n-1); 64 printf("%I64d\n",(ans.mat[0][0]*2+m)%m); 65 } 66 return 0; 67 }