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54 lines
1.3 KiB
54 lines
1.3 KiB
#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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const int N = 25;
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int n, m, mod;
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char s[N];
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int ne[N];
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int a[N][N];
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//矩阵乘法
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void mul(int a[][N], int b[][N], int c[][N]) {
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int t[N][N] = {0};
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for (int i = 0; i < N; i++) {
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for (int j = 0; j < N; j++)
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for (int k = 0; k < N; k++)
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t[i][j] = (t[i][j] + (LL)(a[i][k] * b[k][j]) % mod) % mod;
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}
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memcpy(c, t, sizeof t);
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}
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int main() {
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cin >> n >> m >> mod;
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cin >> s + 1;
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// kmp
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for (int i = 2, j = 0; i <= m; i++) {
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while (j && s[j + 1] != s[i]) j = ne[j];
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if (s[j + 1] == s[i]) j++;
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ne[i] = j;
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}
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// 初始化A[i][j]
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for (int i = 0; i < m; i++)
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for (int c = '0'; c <= '9'; c++) {
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int j = i;
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while (j && s[j + 1] != c) j = ne[j];
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if (s[j + 1] == c) j++;
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if (j < m) a[i][j]++;
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}
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// f[0][0]=1 base case
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int f[N][N] = {1};
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//矩阵快速幂
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for (int i = n; i; i >>= 1) {
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if (i & 1) mul(f, a, f); // f:基底 a:需要计算快速幂的常数矩阵 f:结果存储
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mul(a, a, a); //倍增a
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}
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int res = 0;
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for (int i = 0; i < m; i++) res = (res + f[0][i]) % mod;
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printf("%d\n", res);
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return 0;
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} |