從小明家所在公交站出發有n路公交到公司，現給出每路公交的停站數(不包括起點和終點)，及每次停的時間(一路車在每個站停的時間相同)和發車的間隔，先假定每輛車同時在相對時間0分開始發車，且所有車在相鄰兩個站之間的耗時相同,都為5分鐘。給定小明起床的相對時間(相對0的分鐘數)，請計算他最早到達公司的相對時間。

給定每路車的停站數stops,停站時間period,發車間隔interval及公交路數n，出發時間s。請返回最早到達時間。保證公交路數小於等於500，停站數小於等於50。

class TakeBuses {
public:
    int chooseLine(vector<int> stops, vector<int> period, vector<int> interval, int n, int s) {
        // write code here
        int result = 100000 ;
        for ( int i = 0; i < n; ++ i ) {
            int cost_time = s % interval[i] ;
            int wait_time = cost_time == 0 ? 0 : interval[i] - cost_time ;
            result = min( result, wait_time + ( stops[i] + 1 ) * 5 + period[i] * stops[i] ) ;
        }
        
        return result + s ;
    }
};
 

第二次做：

class TakeBuses {
public:
    int chooseLine(vector<int> stops, vector<int> period, vector<int> interval, int n, int s) {
        // write code here
        int result = 1000000 ;
        for ( int i = 0; i < n; ++ i ) {
            int cost_time = s % interval[i] ;
            int wait_time = cost_time == 0 ? 0 : interval[i] - cost_time ;
            result = min( result, wait_time + ( stops[i] * period[i] ) + ( stops[i] + 1 ) * 5 ) ;
        }
        return result + s ;
    }
};
 

第三次做：

class TakeBuses {
public:
    int chooseLine(vector<int> stops, vector<int> period, vector<int> interval, int n, int s) {
        // write code here
        int result = 99999999 ;
        for ( int i = 0; i < n; ++ i ) {
            int tmp = s % interval[i] ;
            int wait_time ;
            if ( tmp == 0 ) wait_time = 0 ;
            else wait_time = interval[i] - tmp ;
            int total = ( stops[i] * period[i] ) + ( stops[i] + 1 ) * 5 + wait_time ;
            if ( total < result ) result = total ;
        }
        return result + s ;
    }
};