Problem--158A--Codeforces--Next Round

Next Round
time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output

“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.

Input
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
Output the number of participants who advance to the next round.

Examples

input
8 5
10 9 8 7 7 7 5 5
output
6

input
4 2
0 0 0 0
output
0

Note
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.

大致題意:
一共有n個人,成績大於或等於第k個人成績的人可以進入下一輪比賽,輸出一共可以有多少人可以進入下一輪比賽

01這個代碼可以忽略,嘻嘻

#include<stdio.h>
int main()
{
    int n,k,count=0;
    scanf("%d%d",&n,&k);
    int num[n];
    for(int i=0; i<n; i++)
        scanf("%d",&num[i]);
    if(num[k-1]==0)
    {
        for(int i=0; i<n; i++)
        {
            if(num[i]>num[k-1])
                count++;
        }
    }
    else
    {
        for(int i=0; i<n; i++)
        {
            if(num[i]>=num[k-1])
                count++;
        }
    }
    printf("%d\n",count);
    return 0;
}

10 更加簡潔

#include <stdio.h>
int main()
{
    int n,k,num[50],i;
    scanf("%d%d",&n,&k);
    for(i=0; i<n; i++) scanf("%d",&num[i]);
    for(i=0; num[i]>=num[k-1]&&num[i]&&i<n; i++);
    printf("%d",i);
}

11再來看看大神的代碼—桶

int s[100];
int main(i,n,k,x)
{
    scanf("%d%d",&n,&k);
    for(i=0; i<n; ++i)scanf("%d",&x),x?s[x-1]++:0;
    for(i=99,x=0; ~i; --i)x<k?x+=s[i]:0;
    printf("%d",x);
}