快速矩阵幂 csu1597 薛XX后代的IQ

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裸快速矩阵的题，，主要来模板测试

说下这个模板用法，下标都是从0开始

Mat(行数，列数，填充矩阵(可填))

mat_mul(A,B)矩阵乘法

mat_pow(A,b)快速矩阵幂

this->S存放矩阵的数组

#include<cstdio>#include<cmath>#include<cstring>#include<queue>#include<vector>#include<functional>#include<algorithm>using namespace std;typedef long long LL;typedef pair<int, int> PII;const int matMX = 3;const int MX = 200000 + 5;const int INF = 0x3f3f3f3f;LL mod;LL power(LL a, LL b) {    LL ret = 1;    while(b) {        if(b & 1) ret = ret * a % mod;        a = a * a % mod;        b >>= 1;    }    return ret;}struct Mat {    int m, n;    LL S[matMX][matMX];    Mat(int a, int b) {        m = a;        n = b;        memset(S, 0, sizeof(S));    }    Mat(int a, int b, LL w[][matMX]) {        m = a;        n = b;        for(int i = 0; i < m; i++) {            for(int j = 0; j < n; j++) {                S[i][j] = w[i][j];            }        }    }};Mat mat_mul(Mat A, Mat B) {    Mat C(A.m, B.n);    for(int i = 0; i < A.m; i++) {        for(int j = 0; j < B.n; j++) {            for(int k = 0; k < A.n; k++) {                C.S[i][j] = (C.S[i][j] + A.S[i][k] * B.S[k][j]) % mod;            }        }    }    return C;}Mat Blank(int m, int n) {    Mat ret(m, n);    for(int i = 0; i < m; i++) {        ret.S[i][i] = 1;    }    return ret;}Mat mat_pow(Mat A, LL b) {    Mat ret = Blank(A.m, A.n);    while(b) {        if(b & 1) ret = mat_mul(ret, A);        A = mat_mul(A, A);        b >>= 1;    }    return ret;}int main() {    int T;    scanf("%d", &T);    while(T--) {        int X, Y, A, B, N;        scanf("%d%d%d%d%lld%d", &X, &Y, &A, &B, &mod, &N);        LL T1[][matMX] = {{0, 1}, {A, B}};        LL T2[][matMX] = {{X}, {Y}};        Mat s(2, 2, T1);        Mat ans(2, 1, T2);        Mat res = mat_mul(mat_pow(s, N), ans);        printf("%lld\n", res.S[1][0]);    }    return 0;}

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