hdu 4686 Arc of Dream（矩阵快速幂）

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题目链接：hdu 4686 Arc of Dream

代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 10;const int mod = 1e9 + 7;struct Mat {    int r, c;    ll s[maxn][maxn];    Mat(int r = 0, int c = 0): r(r), c(c) { memset(s, 0, sizeof(s)); }    void init(int r, int c) {        this->r = r;        this->c = c;        memset(s, 0, sizeof(s));    }};Mat ans, x, tmp;void mul_mat(Mat& a, Mat& b, Mat& c) {    tmp.init(a.r, b.c);    for (int k = 0; k < a.c; k++) {        for (int i = 0; i < a.r; i++) {            for (int j = 0; j < b.c; j++)                tmp.s[i][j] = (tmp.s[i][j] + a.s[i][k] * b.s[k][j] % mod) % mod;        }    }    c = tmp;}void pow_mod(Mat& ret, Mat& x, ll n) {    while (n) {        if (n&1) mul_mat(x, ret, ret);        mul_mat(x, x, x);        n >>= 1;    }}ll N;int A0, Ax, Ay, B0, Bx, By;int main () {    while (scanf("%lld", &N) == 1) {        scanf("%d%d%d", &A0, &Ax, &Ay);        scanf("%d%d%d", &B0, &Bx, &By);        ans.init(5, 1);        ans.s[0][0] = 1, ans.s[1][0] = A0 % mod, ans.s[2][0] = B0 % mod;        ans.s[3][0] = 1LL * A0 * B0 % mod;        x.init(5, 5);        x.s[0][0] = 1;        x.s[1][0] = Ay % mod, x.s[1][1] = Ax % mod;        x.s[2][0] = By % mod, x.s[2][2] = Bx % mod;        x.s[3][0] = 1LL * Ay * By % mod, x.s[3][1] = 1LL * Ax * By % mod;        x.s[3][2] = 1LL * Bx * Ay % mod, x.s[3][3] = 1LL * Ax * Bx % mod;        x.s[4][3] = x.s[4][4] = 1;        pow_mod(ans, x, N);        printf("%lld\n", ans.s[4][0]);    }    return 0;}/* * 1    0    0    0    0 * Ay   Ax   0    0    0 * By   0    Bx   0    0 * AyBy AxBy BxAy AxBx 0 * 0    0    0    1    1 */

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