'commit'

TangDou/AcWing_TiGao/T5/JuZhengChengFa/1305_JZ_DP.cpp

@@ -0,0 +1,54 @@
+#include <bits/stdc++.h>
+using namespace std;
+typedef long long LL;
+const int N = 25;
+
+int n, m, mod;
+char s[N];
+int ne[N];
+int a[N][N];
+
+//矩阵乘法
+void mul(int a[][N], int b[][N], int c[][N]) {
+    int t[N][N] = {0};
+    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]) % mod) % mod;
+    }
+    memcpy(c, t, sizeof t);
+}
+
+int main() {
+    cin >> n >> m >> mod;
+    cin >> s + 1;
+
+    // kmp
+    for (int i = 2, j = 0; i <= m; i++) {
+        while (j && s[j + 1] != s[i]) j = ne[j];
+        if (s[j + 1] == s[i]) j++;
+        ne[i] = j;
+    }
+
+    // 初始化A[i][j]
+    for (int i = 0; i < m; i++)
+        for (int c = '0'; c <= '9'; c++) {
+            int j = i;
+            while (j && s[j + 1] != c) j = ne[j];
+            if (s[j + 1] == c) j++;
+            if (j < m) a[i][j]++;
+        }
+
+    // f[0][0]=1 base case
+    int f[N][N] = {1};
+    //矩阵快速幂
+    for (int i = n; i; i >>= 1) {
+        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
+        mul(a, a, a);            //倍增a
+    }
+
+    int res = 0;
+    for (int i = 0; i < m; i++) res = (res + f[0][i]) % mod;
+    printf("%d\n", res);
+    return 0;
+}