[poj3735 Training little cat]【矩阵快速幂】

来源：互联网发布：箪食壶浆以迎将军乎编辑：程序博客网时间：2024/05/17 23:00

好题重温。
唯一的坑点是k可能小于0…

#include <cstdio>#include <cstring>#include <algorithm>const int MAX = 101;typedef long long ll;struct mat {    ll a[MAX][MAX];    mat() {        memset(a, 0, sizeof(a));    }    mat(int x) {        memset(a, 0, sizeof(a));        for (int i = 0; i < MAX; ++i) {            a[i][i] = 1;        }    }    mat operator+(const mat& B)const {        mat res;        for (int i = 0; i < MAX; ++i) {            for (int j = 0; j < MAX; ++j) {                res.a[i][j] = a[i][j] + B.a[i][j];            }        }        return res;    }    mat operator*(const mat& B)const {        mat res;        for (int i = 0; i < MAX; ++i) {            for (int j = 0; j < MAX; ++j) {                if (a[i][j]) {                    for (int k = 0; k < MAX; ++k) {                        res.a[i][k] += a[i][j] * B.a[j][k];                    }                }            }        }        return res;    }} E(1);mat fastpow(mat a, int m) {    mat res(1);    while (m) {        if (m & 1) res = res * a;        m >>= 1;        a = a * a;    }    return res;}int main() {    char op;    int c1, c2;    int n, m, k;    while (~scanf(" %d %d %d", &n, &m, &k)) {        if (n == 0 && m == 0 && k == 0) break;        mat cat, b(1);        cat.a[n][0] = 1;        while (k-- > 0) {            scanf(" %c %d", &op, &c1);            if (op == 'g') {                ++b.a[c1-1][n];            } else if (op == 'e') {                memset(b.a[c1-1], 0, sizeof(b.a[c1-1]));            } else {                scanf(" %d", &c2);                std::swap(b.a[c1-1], b.a[c2-1]);            }        }        cat = fastpow(b, m) * cat;        printf("%lld", cat.a[0][0]);        for (int i = 1; i < n; ++i) {            printf(" %lld", cat.a[i][0]);        }        puts("");    }    return 0;}

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