1787. Turn for MEGA

Time limit: 1.0 second
Memory limit: 64 MB
A traffic light at the turn for the “MEGA” shopping center from the Novomoskovskiy highway works in such a way that?k?cars are able to take a turn in one minute. At weekends all the residents of the city drive to the mall to take a shopping, which results in a huge traffic jam at the turn. Administration of the mall ordered to install a camera at the nearby bridge, which is able to calculate the number of cars approaching this turn from the city. The observation started?n?minutes ago. You should use the data from the camera to determine the number of cars currently standing in the traffic jam.

Input

The first line contains integers?k?and?n?(1 ≤?k,?n?≤ 100), which are the number of cars that can take a turn to “MEGA” in one minute and the number of minutes passed from the beginning of observation. The second line contains space-separated integers?a1, …,?an?(0 ≤?ai?≤ 100), where?aiis the number of cars that approached the turn during the?i-th minute. The observation started at morning, when there were no cars at the turn.

Output

Output the number of cars currently standing in the traffic jam.

Samples

input output

5 3 6 7 2	0
5 3 20 0 0	5

題的大概意思就是在一個路口，每分鐘可以有k輛車轉(zhuǎn)頭，n分鐘內(nèi)分別有ai輛車通過，輸出有多少車滯留。
剛開始想的太簡單了，直接用了個sum把ai全加起來和k*n比較，后來發(fā)現(xiàn)下一分鐘接近路口的車不能出現(xiàn)在上一分鐘里。
#include<cstdio> #include<iostream> using namespace std; int main() {int k,n,sum;int a[1001];while(cin>>k>>n){sum=0;for(int i=0;i<n;i++){cin>>a[i];if(sum+a[i]>k)sum+=a[i]-k;elsesum=0;}cout<<sum<<endl;}return 0; }

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