play game, BINARY SEARCH (IMPORTANT)

https://www.lintcode.com/problem/1671/description?_from=cat

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N individuals are playing games, each game has a referee and N-1 civilian players. Given an array A, A[i] represents that the player i needs to be at least a civilian A[i] times, returning the minimum number of games played.

∑Ai<=1e18
1 < n < 1000

Example 1:
Input：A = [2, 2, 2, 2]
Output : 3
Explanation：
A[0] = 2 means that player 0 needs to be at least 2 times civilian
The first game: Player 0 serves as the referee, at this time A[0] = 0, A[1] = 1, A[2] = 1, A[3] =1
Second game: Player 1 serves as the referee, at this time A[0] = 1, A[1] = 1, A[2] = 2, A[3] = 2
The third game: Player 2 serves as the referee, at this time A[0] = 2, A[1] = 2, A[2] = 2, A[3] = 3
At this point, each player has met the requirements, so you can play three games.

Example 2:
Input：A = [84,53]
Output : 137
Explanation：
The first game: Player 1 serves as the referee, at this time A[0] = 1, A[1] = 0
.
.
.
The 31st game: Player 1 serves as the referee, at this time A[0] = 31, A[1] = 0
Thirty-second game: Player 0 serves as the referee, at this time A[1] = 31, A[1] = 1
Thirty-third game: Player 1 serves as the referee, at this time A[1] = 32, A[1] = 1
Thirty-fourth game: Player 0 serves as the referee, at this time A[1] = 32, A[1] = 2
.
.
.
The 137th game: Player 1 serves as the referee, at this time A[1] = 84, A[1] = 53
At this point, each player has met the requirements, so you can play 137 games.

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class Solution {
public:
    bool isOK(vector<int> &A, long long matches){
        long long referee=0;
        for(auto x: A){
            referee+=(matches-x); // matches-x is the number of times this guy will referee
        }
        return referee>=matches;  // there must be aways sufficient amount of refree for matches
    }

    long long playGames(vector<int> &a) {
        long long low=*max_element(a.begin(),a.end());
        long long high=accumulate(a.begin(),a.end(),0LL);
        long long res=0;
        while(low<=high){
            long long mid=(low+high)/2;
            if(isOK(a,mid)==true){
                high=mid-1;
                res=mid;
            }
            else{
                low=mid+1;
            }
        }
        return res;
    }
};