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TangDou/AcWingSuanFaJinJieZhiNan/P3390.cpp

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+#include <bits/stdc++.h>
+using namespace std;
+//矩阵快速幂
+typedef long long LL;
+const int MOD = 1e9 + 7;
+const int N = 110;
+LL n, k; //本题数据范围很大，用int直接wa哭了
+LL C[N][N], A[N][N], B[N][N];
+
+//矩阵乘法
+void mul(LL C[][N], LL A[][N], LL B[][N]) {
+    //为什么非得要开一个临时的tmp来存结果呢，直接用C确实不行，原理不清楚
+    static LL tmp[N][N];
+    memset(tmp, 0, sizeof tmp);
+    for (LL i = 0; i < n; i++)
+        for (LL j = 0; j < n; j++)
+            for (LL k = 0; k < n; k++) {
+                tmp[i][j] += A[i][k] * B[k][j] % MOD; //矩阵乘法
+                tmp[i][j] %= MOD;
+            }
+    memcpy(C, tmp, sizeof tmp);
+}
+
+void qmi() {
+    //将结果矩阵初始化为单位矩阵
+    memset(B, 0, sizeof B);
+    for (int i = 0; i < n; i++) B[i][i] = 1;
+    //其实就是把整数快速幂修改为矩阵快速幂
+    while (k) { //二进制快速幂
+        if (k & 1) mul(B, B, A); // 联想一下整数快速幂
+        mul(A, A, A);            // base 翻倍
+        k >>= 1;
+    }
+}
+int main() {
+    cin >> n >> k;
+    //输入原始矩阵
+    for (int i = 0; i < n; i++)
+        for (int j = 0; j < n; j++)
+            cin >> A[i][j];
+
+    //计算矩阵快速幂
+    qmi();
+
+    //输出
+    for (int i = 0; i < n; i++) {
+        for (int j = 0; j < n; j++)
+            cout << B[i][j] << " ";
+        cout << endl;
+    }
+    return 0;
+}