You are given a list of songs where the i^th song has a duration of time[i] seconds.

Return the number of pairs of songs for which their total duration in seconds is divisible by 60. Formally, we want the number of indices i, j such that i < j with (time[i] + time[j]) % 60 == 0.

Example 1:

Input: time = [30,20,150,100,40]
Output: 3
Explanation: Three pairs have a total duration divisible by 60:
(time[0] = 30, time[2] = 150): total duration 180
(time[1] = 20, time[3] = 100): total duration 120
(time[1] = 20, time[4] = 40): total duration 60

Example 2:

Input: time = [60,60,60]
Output: 3
Explanation: All three pairs have a total duration of 120, which is divisible by 60.

   题目是需要求数组中，一对数字对，满足二者之和能被60整除。这道题很容易可以写出暴力求解的版本。

class Solution {
public:
    int numPairsDivisibleBy60(vector<int>& time) {
          // 暴力法
        int n=time.size();
        int res=0;
        for(int i=0;i<n;++i){
            for(int j=i+1;j<n;++j){
                if((time[i]+time[j])%60==0) res++;
            }
        }
        return res;
    }
};

    再数字较多的情况下就会超时，那么该如何优化呢？ 暴力法的之间复杂是o(n*n), 这时候再回过头看题目的要求，(time[i]+time[j])%==0 <=>time[i]%60+time[j]%60==0, 此时有两种情况满足上述要求，要么两个数字time[i],time[j]都能被60整除，要么其余数加和等于60，这个时候用一个mark数组记录当前余数的个数，并同时更新结果就好了。这种解法时间复杂度为o(n),可以说非常的高效了。

    感觉这种优化思路非常有意思。

class Solution {
public:
    int numPairsDivisibleBy60(vector<int>& time) {
          // 暴力法
        int res=0;
        int mark[60]={0};
        for(int i=0;i<time.size();++i){
            int a=time[i]%60;
            if(a==0){
                res+=mark[0]; //查看之前该种余数的个数
            }
            else{
                res+=mark[60-a];
            }
            mark[a]++; //要之后更新
        }
        return res;
    }
};