您的位置:首页 > 其它

hdu 5435 Peace small elephant(矩阵快速幂)

2015-09-12 22:15 211 查看
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>
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: