HDU4565(SummerTrainingDay05-C 矩阵快速幂)
So Easy!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4673 Accepted Submission(s): 1539
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
根据条件可证(a-√b)n小于1，(a+√b)n向上取整即为求(a+√b)n + (a-√b)n，问题又转换为a^n+b^n的形式。
1 //2017-08-05 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 #include <algorithm> 6 #define ll long long 7 8 using namespace std; 9 10 ll a, b, n; 11 ll p, q; 12 const int N = 5; 13 ll MOD; 14 15 struct Matrix 16 { 17 ll a[N][N]; 18 int r, c; 19 }ori, res; 20 21 void init() 22 { 23 memset(res.a, 0, sizeof(res.a)); 24 res.r = 1; res.c = 2; 25 res.a[1][1] = p; 26 res.a[1][2] = 2; 27 ori.r = 2; ori.c = 2; 28 ori.a[1][1] = p; 29 ori.a[1][2] = 1; 30 ori.a[2][1] = -q; 31 ori.a[2][2] = 0; 32 } 33 34 Matrix multi(Matrix x, Matrix y) 35 { 36 Matrix z; 37 memset(z.a, 0, sizeof(z.a)); 38 z.r = x.r, z.c = y.c; 39 for(int i = 1; i <= x.r; i++) 40 { 41 for(int k = 1; k <= x.c; k++) 42 { 43 if(x.a[i][k] == 0) continue; 44 for(int j = 1; j<= y.c; j++) 45 z.a[i][j] = (z.a[i][j] + (x.a[i][k] * y.a[k][j]) % MOD + MOD) % MOD; 46 } 47 } 48 return z; 49 } 50 51 void Matrix_pow(int n) 52 { 53 while(n) 54 { 55 if(n & 1) 56 res = multi(res, ori); 57 ori = multi(ori, ori); 58 n >>= 1; 59 } 60 printf("%lld\n", res.a[1][1] % MOD); 61 } 62 63 int main() 64 { 65 while(scanf("%lld%lld%lld%lld", &a, &b, &n, &MOD)!=EOF){ 66 p = 2*a; 67 q = a*a-b; 68 init(); 69 if(n == 0)printf("2\n"); 70 else if(n == 1)printf("%lld\n", p); 71 else Matrix_pow(n-1); 72 } 73 74 return 0; 75 }
Source
