2 problem

Author: hitzzc

README.md

@@ -226,7 +226,9 @@ Golang solution for leetcode. For each problem, there is a simple *_test.go to t
 #### [306. Additive Number](https://github.com/hitzzc/go-leetcode/tree/master/additive_number)
 #### [307. Range Sum Query - Mutable](https://github.com/hitzzc/go-leetcode/tree/master/range_sum_query_mutable)
 #### [309. Best Time to Buy and Sell Stock with Cooldown](https://github.com/hitzzc/go-leetcode/tree/master/best_time_to_buy_and_sell_stock_with_cooldown)
-#### [310. Minimum Height Trees](https://github.com/hitzzc/go-leetcode/tree/master/minimum_height_trees)
+#### [312. Burst Balloons](https://github.com/hitzzc/go-leetcode/tree/master/burst_balloons)
+#### [313. Super Ugly Number](https://github.com/hitzzc/go-leetcode/tree/master/super_ugly_number)
+
 
 
 

burst_balloons/burst_balloons.cpp

@@ -0,0 +1,17 @@
+class Solution {
+public:
+    int maxCoins(vector<int>& nums) {
+        nums.insert(nums.begin(), 1);
+        nums.push_back(1);
+        vector<vector<int>> matrix(nums.size(), vector<int>(nums.size(), 0));
+        for (int k = 2; k < nums.size(); ++k) {
+        	for (int low = 0; low < nums.size()-k; ++low) {
+        		int high = low+k;
+        		for (int i = low+1; i < high; ++i) {
+        			matrix[low][high] = max(matrix[low][high], nums[low]*nums[i]*nums[high] + matrix[low][i] + matrix[i][high]);
+        		}
+        	}
+        }
+        return matrix[0][matrix.size()-1];
+    }
+};

super_ugly_number/super_ugly_number.cpp

@@ -0,0 +1,17 @@
+class Solution {
+public:
+    int nthSuperUglyNumber(int n, vector<int>& primes) {
+        vector<int> dp(n, 1);
+        vector<int> idxs(primes.size(), 0);
+        for (int i = 1; i < n; ++i) {
+        	dp[i] = INT_MAX;
+        	for (int idx = 0; idx < idxs.size(); ++idx) {
+        			dp[i] = min(dp[i], primes[idx]*dp[idxs[idx]]);
+        	}
+        	for (int idx = 0; idx < idxs.size(); ++idx) {
+        		if (dp[i] == primes[idx] * dp[idxs[idx]]) ++idxs[idx];
+        	}
+        }
+        return dp.back();
+    }
+};