Minimum Time to Complete Trips
You are given an array time where time[i] denotes the time taken by the ith bus to complete one trip.
Each bus can make multiple trips successively; that is, the next trip can start immediately after completing the current trip. Also, each bus operates independently; that is, the trips of one bus do not influence the trips of any other bus.
You are also given an integer totalTrips, which denotes the number of trips all buses should make in total.
Return the minimum time required for all buses to complete at least totalTrips trips.
Example 1:
Input: time = [1,2,3], totalTrips = 5 Output: 3 Explanation: - At time t = 1, the number of trips completed by each bus are [1,0,0]. The total number of trips completed is 1 + 0 + 0 = 1. - At time t = 2, the number of trips completed by each bus are [2,1,0]. The total number of trips completed is 2 + 1 + 0 = 3. - At time t = 3, the number of trips completed by each bus are [3,1,1]. The total number of trips completed is 3 + 1 + 1 = 5. So the minimum time needed for all buses to complete at least 5 trips is 3.
Example 2:
Input: time = [2], totalTrips = 1 Output: 2 Explanation: There is only one bus, and it will complete its first trip at t = 2. So the minimum time needed to complete 1 trip is 2.
Constraints:
1 <= time.length <= 10^51 <= time[i], totalTrips <= 10^7
题解
经过对题目的分析可知，题目要求一个最小阈值，这个最小的阈值能够满足下面的条件：
代码
class Solution {
public:
int n,c;
vector<int>a;
bool isok(long long s){
long long sum=0;
for(int i=0;i<n;i++){
sum+=s/a[i];
}
return sum>=c;
}
long long minimumTime(vector<int>& time, int totalTrips) {
n=time.size();
for(int i=0;i<n;i++){
a.push_back(time[i]);
}
c=totalTrips;
long long MAX=0;
for(int i=0;i<n;i++){
MAX=max(MAX,(long long)time[i]);
}
MAX*=totalTrips;
long long l=0,r=MAX,mid,ans;
while(l<=r){
mid=l+(r-l)/2;
if(isok(mid))
ans=mid,r=mid-1;
else
l=mid+1;
}
return ans;
}
};
扫描二维码关注公众号，回复：
16843273 查看本文章
