codeforces 158A Next Round (水题)
来源:互联网 发布:合伙人 知乎 编辑:程序博客网 时间:2024/06/05 15:54
题目:
"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.
8 510 9 8 7 7 7 5 5
6
4 20 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.
题意分析:
代码:
#include<iostream>int n,k,i,j,a[51];using namespace std;int main(){ cin>>n>>k; while(n>i) cin>>a[i++]; while(a[j]&&a[j]>=a[k-1])++j; cout<<j;}
- codeforces 158A Next Round(水题)
- codeforces 158A Next Round (水题)
- CodeForces 158 A.Next Round(水~)
- CodeForces 158A Next Round
- codeforces 158A Next Round
- Problem--158A--Codeforces--Next Round
- CodeForces.158A Next Round (水模拟)
- 158A - Next Round
- 158A Next Round
- Codeforces VK Cup 2012 Qualification Round 1 / 158A Next Round(模拟)
- CF#158A Next Round
- code Forces 158A Next Round
- A. Next Round
- CodeForces 659A Round House(水题)
- Codeforces Round #182 (Div. 2)A(水题)
- Codeforces Round #354 (Div. 2) A (水题)
- Educational Codeforces Round 13 A Johny Likes Numbers(水题)
- Codeforces Round #360 (Div. 2) A Opponents(水题)
- redis缓存的安装和使用
- Ubuntu下搭建TQ2440的程序下载环境
- 使用JSON的方法
- Linux文件系统的目录结构
- 第二十四周工作日志
- codeforces 158A Next Round (水题)
- 汇编语言04——[BX]和loop指令
- hdu 3221 Brute-force Algorithm(快速幂取模,矩阵快速幂求fib)
- 分公司电脑通过大手笔投入
- Linux文件类型与扩展名
- 快上课时间和高科技会所开个会看个够
- if中逻辑运算符表达式运行顺序
- tomcat 7.0的虚拟目录配置方法
- 再解 KMP(初学)