479. Largest Palindrome Product

https://leetcode.com/problems/largest-palindrome-product/description/
    long getCandidate(int n) {
        string s = to_string(n);
        reverse(s.begin(), s.end());
        return stol(to_string(n) + s);
    }
    int largestPalindrome(int n) {
        if(n==1) return 9;
        int mx = pow(10,n) - 1;
        int mi = pow(10,n-1);
        for(int i=mx; i>=mi; i--) {
            long c = getCandidate(i);
            for(long j=mx; j*j>=c; j--) {
                if(c%j == 0 && c/j >= mi) {
                    return c%1337;
                }
            }
        }
        return -1;
    }

278. First Bad Version

https://leetcode.com/problems/first-bad-version/description/
    int firstBadVersion(int n) {
        int lo = 1, hi = n;
        while(lo<hi){
            int mid = lo +(hi - lo)/2;
            if(isBadVersion(mid)) hi = mid;
            else lo = mid + 1;
        }
        return lo;
    }

374. Guess Number Higher or Lower

https://leetcode.com/problems/guess-number-higher-or-lower/description/
Solution 1. Recursion.
    int guessing(int lo, int hi) {
        int mi = (hi - lo)/2 + lo;
        int c = guess(mi);
        if(c==0) return mi;
        else if(c==1) return guessing(mi+1, hi);
        else return guessing(lo, mi-1);
    }
    int guessNumber(int n) {
        return guessing(1, n);
    }
Solution 2. Iteration.
    int guessNumber(int n) {
        int mi, re, hi = n, lo = 1;
        while(true) {
            mi = (hi - lo)/2 + lo;
            re = guess(mi);
            if(re == 0) return mi;
            else if(re == 1) lo = mi + 1;
            else hi = mi - 1;
        }
    }

172. Factorial Trailing Zeroes

https://leetcode.com/problems/factorial-trailing-zeroes/description/
Solution. The number of trailing 0 of n! equals the total number of factor 5 in 1, 2, ... , n. 
Each step, n/5 is the number of integers that have at least one factor of 5 in 1,2,...,n.
    int trailingZeroes(int n) {
        if(n == 0) return 0;
        int nz = 0;
        while(n) {
            nz += n/5;
            n /= 5;
        }
        return nz;
    }