hdu 5435 Peace small elephant（矩阵快速幂）

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题目链接：hdu 5435 Peace small elephant

宽度很小，长很大，明显是矩阵快速幂，枚举两个二进制状态，判断一下是否可以转移，建出矩阵。

<span style="font-size:18px;">#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;typedef long long ll;const int maxn = (1<<7) + 5;const int mod = 1000000007;struct Mat {int r, c; ll s[maxn][maxn];void init(int r = 0, int c = 0) {this->r = r;this->c = c;memset(s, 0, sizeof(s));}}X[10];int N, M;Mat ans;Mat mul(Mat& a, Mat& b) {Mat ret;ret.init(a.r, b.c);for (int k = 0; k < a.c; k++) {for (int i = 0; i < a.r; i++) {if (a.s[i][k] == 0) continue;for (int j = 0; j < b.c; j++)ret.s[i][j] = (ret.s[i][j] + a.s[i][k] * b.s[k][j]) % mod;}}return ret;}void powMat (Mat& ret, Mat x, int n) {while (n) {if (n&1) ret = mul(x, ret);x = mul(x, x);n >>= 1;}}bool judge (int s, int t, int n) {int k = 0;for (int i = 0; i < n; i++) {int f = ((s>>i)&1)*2 + ((t>>i)&1);if ((k&f) == 0 && k != 0 && f != 0) return false;k = f;}return true;}void init() {for (int k = 1; k <= 7; k++) {int n = (1<<k);X[k].init(n, n);for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {if (judge(i, j, k))X[k].s[i][j] = 1;}}}}int main () {init();while (scanf("%d%d", &N, &M) == 2) {ans.init((1<<M), 1);for (int i = 0; i < (1<<M); i++) ans.s[i][0] = 1;powMat(ans, X[M], N-1);ll ret = 0;for (int i = 0; i < (1<<M); i++) ret = (ret + ans.s[i][0]) % mod;printf("%d\n", (int)ret);}return 0;}</span>

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