1 介绍

本博客用来记录同余和矩阵乘法相关的题目。

2 训练

题目1：203同余方程

C++代码如下，

#include <iostream>
#include <algorithm>
#include <cstring>using namespace std;typedef long long LL;int exgcd(int a, int b, int &x, int &y) {if (!b) {x = 1, y = 0;return a;}int d = exgcd(b, a % b, y, x);y -= a / b * x;return d;
}int main() {int a, b;cin >> a >> b;int x, y;exgcd(a, b, x, y);cout << (x % b + (LL)b) % b << endl;return 0;
}

题目2：222青蛙的约会

C++代码如下，

#include <cstring>
#include <iostream>
#include <algorithm>using namespace std;typedef long long LL;LL exgcd(LL a, LL b, LL &x, LL &y) {if (!b) {x = 1, y = 0;return a;}LL d = exgcd(b, a % b, y, x);y -= a / b * x;return d;
}int main() {LL a, b, m, n, L;cin >> a >> b >> m >> n >> L;LL x, y;LL d = exgcd(m - n, L, x, y);if ((b - a) % d) puts("Impossible");else {x *= (b - a) / d;LL t = abs(L / d);cout << (x % t + t) % t << endl;}return 0;
}

题目3：202最幸运的数字

C++代码如下，

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>using namespace std;typedef long long LL;LL qmul(LL a, LL k, LL b) {LL res = 0;while (k) {if (k & 1) res = (res + a) % b;a = (a + a) % b;k >>= 1;}return res;
}LL qmi(LL a, LL k, LL b) {LL res = 1;while (k) {if (k & 1) res = qmul(res, a, b);a = qmul(a, a, b);k >>= 1;}return res;
}LL get_euler(LL C) {LL res = C;for (LL i = 2; i <= C / i; ++i) {if (C % i == 0) {while (C % i == 0) C /= i;res = res / i * (i - 1);}}if (C > 1) res = res / C * (C - 1);return res;
}int main() {int T = 1;LL L;while (cin >> L, L) {int d = 1;while (L % (d * 2) == 0 && d * 2 <= 8) d *= 2;LL C = 9 * L / d;LL phi = get_euler(C);LL res = 1e18;if (C % 2 == 0 || C % 5 == 0) res = 0;else {for (LL d = 1; d * d <= phi; d++) {if (phi % d == 0) {if (qmi(10, d, C) == 1) res = min(res, d);if (qmi(10, phi / d, C) == 1) res = min(res, phi / d);}}}printf("Case %d: %lld\n", T++, res);}return 0;
}

题目4：1298曹冲养猪

C++代码如下，

#include <cstring>
#include <iostream>
#include <algorithm>using namespace std;typedef long long LL;const int N = 10;int n;
int A[N], B[N];void exgcd(LL a, LL b, LL &x, LL &y) {if (!b) x = 1, y = 0;else {exgcd(b, a % b, y, x);y -= a / b * x;}
}int main() {scanf("%d", &n);LL M = 1;for (int i = 0; i < n; ++i) {scanf("%d%d", &A[i], &B[i]);M *= A[i];}LL res = 0;for (int i = 0; i < n; ++i) {LL Mi = M / A[i];LL ti, x;exgcd(Mi, A[i], ti, x);res += B[i] * Mi * ti;}cout << (res % M + M) % M << endl;return 0;
}

题目5：1303斐波那契前 n 项和

C++代码如下，

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>using namespace std;typedef long long LL;const int N = 3;int n, m;void mul(int c[], int a[], int b[][N]) {int temp[N] = {0};for (int i = 0; i < N; ++i) {for (int j = 0; j < N; ++j) {temp[i] = (temp[i] + (LL)a[j] * b[j][i]) % m;}}memcpy(c, temp, sizeof temp);
}void mul(int c[][N], int a[][N], int b[][N]) {int temp[N][N] = {0};for (int i = 0; i < N; ++i) {for (int j = 0; j < N; ++j) {for (int k = 0; k < N; ++k) {temp[i][j] = (temp[i][j] + (LL)a[i][k] * b[k][j]) % m;}}}memcpy(c, temp, sizeof temp);
}int main() {cin >> n >> m;int f1[N] = {1, 1, 1};int a[N][N] = {{0, 1, 0},{1, 1, 1},{0, 0, 1}};n--;while (n) {if (n & 1) mul(f1, f1, a);mul(a, a, a);n >>= 1;}cout << f1[2] % m << endl;return 0;
}

题目6：1304佳佳的斐波那契

C++代码如下，

#include <cstring>
#include <iostream>
#include <algorithm>using namespace std;typedef long long LL;const int N = 4;int n, m;void mul(int c[][N], int a[][N], int b[][N]) {static int t[N][N];memset(t, 0, sizeof t);for (int i = 0; i < N; ++i) {for (int j = 0; j < N; ++j) {for (int k = 0; k < N; ++k) {t[i][j] = (t[i][j] + (LL)a[i][k] * b[k][j]) % m;}}}memcpy(c, t, sizeof t);
}int main() {cin >> n >> m;int f1[N][N] = {1, 1, 1, 0};int a[N][N] = {{0, 1, 0, 0},{1, 1, 1, 0},{0, 0, 1, 1},{0, 0, 0, 1}};int k = n - 1;while (k) {if (k & 1) mul(f1, f1, a);mul(a, a, a);k >>= 1;}cout << (((LL)n * f1[0][2] - f1[0][3]) % m + m) % m << endl;return 0;
}

题目7：1305GT考试

C++代码如下，

#include <cstring>
#include <iostream>
#include <algorithm>using namespace std;const int N = 25;int n, m, mod;
char str[N];
int ne[N];
int a[N][N];void mul(int c[][N], int a[][N], int b[][N]) {static int t[N][N];memset(t, 0, sizeof t);for (int i = 0; i < m; ++i) {for (int j = 0; j < m; ++j) {for (int k = 0; k < m; ++k) {t[i][j] = (t[i][j] + a[i][k] * b[k][j]) % mod;}}}memcpy(c, t, sizeof t);
}int qmi(int k) {int f0[N][N] = {1};while (k) {if (k & 1) mul(f0, f0, a);mul(a, a, a);k >>= 1;}int res = 0;for (int i = 0; i < m; ++i) res = (res + f0[0][i]) % mod;return res;
}int main() {cin >> n >> m >> mod;cin >> str + 1;for (int i = 2, j = 0; i <= m; ++i) {while (j && str[j + 1] != str[i]) j = ne[j];if (str[j+1] == str[i]) j++;ne[i] = j;}for (int j = 0; j < m; ++j) {for (int c = '0'; c <= '9'; ++c) {int k = j;while (k && str[k+1] != c) k = ne[k];if (str[k+1] == c) k++;if (k < m) a[j][k]++;}}cout << qmi(n) << endl;return 0;
}