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UVA 10689 Yet another Number Sequence 矩陣快速冪 水呀水

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技術分享

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;
const int N = 4;
int Mod;
int msize;

struct Mat
{
    int mat[N][N];
};

Mat operator *(Mat a, Mat b)
{
    Mat c;
    memset(c.mat, 
0, sizeof(c.mat)); for(int k = 0; k < msize; ++k) for(int i = 0; i < msize; ++i) if(a.mat[i][k]) for(int j = 0; j < msize; ++j) if(b.mat[k][j]) c.mat[i][j] = ((ll)a.mat[i][k] * b.mat[k][j] + c.mat[i][j])%Mod;
return c; } Mat operator ^(Mat a, int k) { Mat c; memset(c.mat,0,sizeof(c.mat)); for(int i = 0; i < msize; ++i) c.mat[i][i]=1; for(; k; k >>= 1) { if(k&1) c = c*a; a = a*a; } return c; } int main() { // freopen("in.txt", "r", stdin);
int t, a, b, n, m; msize = 2; scanf("%d", &t); while(t--) { scanf("%d%d%d%d", &a, &b, &n, &m); Mod = 1; while(m--) Mod *= 10; if(n == 0) { printf("%d\n", a%Mod); continue; } Mat A; A.mat[0][0] = 1, A.mat[0][1] = 1; A.mat[1][0] = 1, A.mat[1][1] = 0; A = A^(n-1); printf("%d\n", (A.mat[0][0]*b + A.mat[0][1]*a)%Mod); } return 0; }

UVA 10689 Yet another Number Sequence 矩陣快速冪 水呀水