Tokyo! Even though the trip was very rushed, it seemed to be enough for me. Even if I had extra time, I'd probably just roam around, see a few more landmarks and buildings, and enjoy more of the renowned local delicacies. Perhaps for every traveler, every city has its own advantages to admire.

https://leetcode.com/problems/count-all-valid-pickup-and-delivery-options/ https://leetcode.com/problems/count-all-valid-pickup-and-delivery-options/submissions/1045563113/ class Solution { public: int countOrders(int n) { vector<int> dp(n + 1); dp[1] = 1; for(int i = 2; i <= n; ++i){ int tn = (2 * i - 1); int sn = (tn * (tn + 1)) / 2; dp[i] = (long long)dp[i - 1] * sn % 1000000007; } return dp[n]; } }; Looked Hint

https://leetcode.com/problems/extra-characters-in-a-string/ https://leetcode.com/problems/extra-characters-in-a-string/submissions/1038162877/ class Solution { public: vector<int> ans; unordered_map<string, bool> mp; int minExtraChar(string s, vector<string>& dictionary) { ans.resize(s.size()+1, 0x3f3f3f3f); for(string a : dictionary) mp[a] = true; ans[s.size() - 1] = !mp[string(1, s.back())]; ans[s.size()] = 0; for(int i = s.size() - 2; i >= 0; --i){ for(int j = 1; j <= s.size() - i; ++j) { string tmp = s.substr(i, j); ans[i] = min(ans[i], (mp[tmp]?0:j) + ans[i+j]); } } return ans[0]; } };

https://leetcode.com/problems/minimum-number-of-taps-to-open-to-water-a-garden/ https://leetcode.com/problems/minimum-number-of-taps-to-open-to-water-a-garden/submissions/1036407008/ class Solution { public: int minTaps(int n, vector<int>& ranges) { vector<int> dp(n + 1, 0x3f3f3f3f); dp[0] = 0; for(int i = 0; i < ranges.size(); ++i) { int l = max(0, i - ranges[i]); int r = min(n, i + ranges[i]); for(int j = l; j <= r; ++j) { dp[j] = min(dp[j], dp[l] + 1); } } int ans = 0; for(int i = 0; i <= n; ++i) { if(dp[i] == 0x3f3f3f3f) { return -1; } ans = max(ans, dp[i]); } return ans; } };

https://leetcode.com/problems/minimum-replacements-to-sort-the-array/description/ https://leetcode.com/problems/minimum-replacements-to-sort-the-array/submissions/1035547342/ class Solution { public: long long minimumReplacement(vector<int>& nums) { long long ans = 0; int Max = nums.back(); for(int i = nums.size() - 2; i >= 0; --i) { //cout << i << " max = " << Max << " num = " << nums[i] << endl; if(nums[i] <= Max) { Max = nums[i]; continue; } int divisor = nums[i]; int dividend = Max; int mod = divisor % dividend; int quotient = divisor / dividend; if(mod == 0) { ans += quotient - 1; continue; } ans += quotient; // binary search to find max Max int l = mod, r = Max; while(l <= r) { int mid = (l + r) >> 1; int gap = mid - mod; int times = gap / quotient; if(gap % quotient !

cstdlib div div_t div(int numer, int denom); ldiv_t div(long int numer, long int denom); lldiv_t div (long long int numer, long long int denom); Returns the integral quotient and remainder of the division of numer by denom ( numer/denom ) as a structure of type div_t, ldiv_t or lldiv_t, which has two members: quot and rem. Only int version for c /* div example */ #include <stdio.h> /* printf */ #include <stdlib.