hihocode xiaohi

HihoCode/小h的朋友们.cpp

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+#include <stdio.h>
+#include <stdlib.h>
+#include <vector>
+#include <iostream>
+#include <map>
+#include <set>
+#include <queue>
+#include <cmath>
+
+using namespace std;
+
+typedef struct Matrix
+{
+    long long ma[15][15];
+}Matrix;
+
+Matrix A;   // 矩阵A
+Matrix B;   // 保存最后的结果
+Matrix unit;// 单位矩阵，用于在下面的fun函数中
+
+const int MOD = (1e9 + 7);
+int n = 2;
+// 两个矩阵相乘
+Matrix Mul(Matrix m1, Matrix m2)
+{
+    Matrix c;
+    for(int i=0; i<n; ++i)
+        for(int j=0; j<n; ++j)
+        {
+            c.ma[i][j] = 0;
+            for(int k=0; k<n; ++k)
+            {
+                c.ma[i][j] += m1.ma[i][k] % MOD * m2.ma[k][j] % MOD;
+                c.ma[i][j] %= MOD;
+            }
+        }
+    return c;
+}
+
+// 二分法求矩阵的幂
+void fun(long long num)
+{
+    Matrix in = A;
+    Matrix un = unit;
+
+    while(num > 1)
+    {
+        if(num & 1)  
+        {
+            --num;
+            un = Mul(in, un);   
+        }
+        else
+        {
+            num >>= 1;
+            in = Mul(in, in);
+
+        }
+    }
+    B = Mul(in, un);
+}
+
+long long pm (long long x, long long y) {
+    x = (x % MOD + MOD) % MOD;
+    long long r = 1;
+    while (y) {
+        if (y & 1) (r *= x) %= MOD;
+        (x *= x) %= MOD;
+        y >>= 1;
+    }
+    return r;
+}
+
+int main() {
+    unit.ma[0][0] = 1;
+    unit.ma[1][1] = 1;
+    int k1, k2;
+    while (cin >> k1 >> k2) {
+
+        long long x, y, Ax, Ay;
+        cin >> x >> y >> Ax >> Ay;
+        long long z;
+        cin >> z;
+        if (k1 == 0 && k2 == 0) {
+            cout << 0 << endl;
+            return 0;
+        }
+        if (k2 == 0) {
+            if (x <= z) cout << Ax * pm (k1, z - x) % MOD << endl;
+            else cout << Ax * pm (pm (k1, MOD - 2), x - z) % MOD << endl;
+            return 0;
+        }
+        if (k1 == 0) {
+            if ((x & 1) == (z & 1)) {
+                if (x <= z) cout << Ax * pm (k2, (z - x) / 2) % MOD << endl;
+                else cout << Ax * pm (pm (k2, MOD - 2), (x - z) / 2) % MOD << endl;
+            } else {
+                if (y <= z) cout << Ay * pm (k2, (z - y) / 2) % MOD << endl;
+                else cout << Ay * pm (pm (k2, MOD - 2), (y - z) / 2) % MOD << endl;
+            }
+            return 0;
+        }
+        A.ma[0][0] = k1;
+        A.ma[0][1] = k2;
+        A.ma[1][0] = 1;
+        A.ma[1][1] = 0;
+        fun(x-2);
+        long long X = B.ma[0][0];
+        long long Y = B.ma[0][1]; //Ax
+        fun(y-2);
+        long long XX = B.ma[0][0];
+        long long YY = B.ma[0][1]; //Bx
+        fun(z-2);
+        long long e = B.ma[0][0];
+        long long f = B.ma[0][1];
+        long long yyy = (Ax * XX%MOD - Ay * X %MOD+ MOD)%MOD * pm(Y * XX - X * YY, MOD-2)%MOD;
+        long long xxx = (Ax * YY%MOD - Ay * Y%MOD + MOD)%MOD * pm(X * YY - Y * XX, MOD-2)%MOD;
+        cout << (e * xxx%MOD + f * yyy%MOD) % MOD << endl;
+    }
+}
+
+