372

Author: yubinbai

README.md

@@ -3,6 +3,7 @@ Leetcode solutions in C++, Java, Python
 
 |#|Title|Solution|Difficulty|
 |---|-----|--------|----------|
+|372|[Super Pow](./src/super_pow)|[C++](./src/super_pow)|Medium|
 |371|[Sum of Two Integers](./src/sum_of_two_integers)|[C++](./src/sum_of_two_integers)|Easy|
 |370|[Range Addition](./src/range_addition)|[C++](./src/range_addition)|Medium|
 |369|[Plus One Linked List](./src/plus_one_linked_list)|[C++](./src/plus_one_linked_list)|Medium|

src/sum_of_two_integers/README.md

@@ -1,6 +1,6 @@
 Problem
 ===
-late the sum of two integers a and b, but you are not allowed to use the operator + and -.
+calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.
 
 Example:
 Given a = 1 and b = 2, return 3.

src/super_pow/Makefile

@@ -0,0 +1,6 @@
+all:
+	clang++ main.cpp -o main.o -std=c++11
+
+run:
+	clang++ main.cpp -o main.o -std=c++11
+	./main.o

src/super_pow/README.md

@@ -0,0 +1,17 @@
+Problem
+===
+
+Your task is to calculate a ** b mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array.
+
+Example1:
+
+a = 2
+b = [3]
+
+Result: 8
+Example2:
+
+a = 2
+b = [1,0]
+
+Result: 1024

src/super_pow/main.cpp

@@ -0,0 +1,65 @@
+#include <algorithm>
+#include <cstdlib>
+#include <iostream>
+#include <map>
+#include <queue>
+#include <set>
+#include <sstream>
+#include <stack>
+#include <string>
+#include <unordered_map>
+#include <unordered_set>
+#include <vector>
+
+using namespace std;
+
+class Solution {
+private:
+  int mod = 1337;
+public:
+  int superPow(int a, vector<int>& b) {
+    int n = b.size();
+    int result = 1;
+    for (int i = n - 1; i >= 0; i--) {
+      result = result * mod_pow(a, b[i]) % mod;
+      a = mod_pow(a, 10);
+    }
+    return result;
+  }
+  inline int mod_pow(int a, int b) {
+    int ans = 1;
+    a %= mod;
+    while (b > 0) {
+      if (b & 1) {
+        ans = ans * a % mod;
+      }
+      a = a * a % mod;
+      b >>= 1;
+    }
+    return ans;
+  }
+};
+
+int main() {
+  Solution sol;
+  int a;
+  vector<int> b;
+
+  a = 2;
+  b = {3};
+  cout << sol.superPow(a, b) << endl;
+
+  a = 2;
+  b = {1, 0};
+  cout << sol.superPow(a, b) << endl;
+
+  a = 647;
+  b = {2, 0, 0};
+  cout << sol.superPow(a, b) << endl;
+
+  a = 2147483647;
+  b = {2, 0, 0};
+  cout << sol.superPow(a, b) << endl;
+
+  return 0;
+}