"Contestant who earns a score equal to or greater than the k-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules.
A total of n participants took part in the contest (n ≥ k), and you already know their scores. Calculate how many participants will advance to the next round.
The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 50) separated by a single space.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 100), where ai is the score earned by the participant who got the i-th place. The given sequence is non-increasing (that is, for all i from 1 to n - 1 the following condition is fulfilled: ai ≥ ai + 1).
Output the number of participants who advance to the next round.
8 5
10 9 8 7 7 7 5 5
6
4 2
0 0 0 0
0
In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers.
In the second example nobody got a positive score.
题意:长度为n的有序降序序列,成绩比第k名高或者相等的且不为0的人数
#include<iostream> #include<string> #include<cstring> using namespace std; typedef long long ll; int a[1100]; int main() { int n,k; cin>>n>>k; for(int i=1;i<=n;i++) cin>>a[i]; int cnt=k; for(int i=k+1;i<=n;i++) { if(a[i]>=a[k]) cnt++; } int tmp=cnt; for(int i=1;i<=tmp;i++) if(a[i]==0) cnt--; cout<<cnt; return 0; }