Yet another Number Sequence
Let’s define another number sequence, given by the following function:
f(0) = a
f(1) = b
f(n) = f(n − 1) + f(n − 2), n > 1
When a = 0 and b = 1, this sequence gives the Fibonacci Sequence. Changing the values of a and
b, you can get many different sequences. Given the values of a, b, you have to find the last m digits of
f(n).
Input
The first line gives the number of test cases, which is less than 10001. Each test case consists of a
single line containing the integers a b n m. The values of a and b range in [0,100], value of n ranges in
[0,1000000000] and value of m ranges in [1,4].
Output
For each test case, print the last m digits of f(n). However, you should NOT print any leading zero.
Sample Input
4
0 1 11 3
0 1 42 4
0 1 22 4
0 1 21 4
Sample Output
89
4296
7711
946
题意:
给你 f[0],f[1] 分别为A,B求F[n] % (10^m)
题解:
n有点大,矩阵快速幂
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> using namespace std ; typedef long long ll; const int N = 100000 + 10; const int mod = 1e9 + 7; const int M[55] = {1, 10, 100, 1000, 10000}; struct Matrix { ll mat[2][2]; }U,F,L; ll MOD; Matrix multi (Matrix a, Matrix b) { Matrix ans; for(int i = 0; i < 2; i++) { for(int j = 0; j < 2; j++) { ans.mat[i][j] = 0; for(int k = 0; k < 2; k++) ans.mat[i][j] += a.mat[i][k] * b.mat[k][j]; ans.mat[i][j] %= MOD; } } return ans; } ll a,b,m; Matrix powss(ll n) { Matrix ans = L,p = U; while(n) { if(n&1) ans = multi(p,ans); n >>= 1; p = multi(p,p); } return ans; } int main() { int T; scanf("%d",&T); while(T--) { ll n; scanf("%lld%lld%lld%lld",&a,&b,&n,&m); U = {1,1,1,0}; L = {b,0,a,0}; MOD = M[m]; Matrix ans = powss(n); printf("%lld ",ans.mat[1][0]); } return 0; }