CodeForces
来源:互联网 发布:b2b数据采购平台 编辑:程序博客网 时间:2024/06/05 15:32
n people are standing in a line to play table tennis. At first, the first two players in the line play a game. Then the loser goes to the end of the line, and the winner plays with the next person from the line, and so on. They play until someone wins k games in a row. This player becomes the winner.
For each of the participants, you know the power to play table tennis, and for all players these values are different. In a game the player with greater power always wins. Determine who will be the winner.
The first line contains two integers: n and k (2 ≤ n ≤ 500, 2 ≤ k ≤ 1012) — the number of people and the number of wins.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ n) — powers of the player. It's guaranteed that this line contains a valid permutation, i.e. all ai are distinct.
Output a single integer — power of the winner.
2 21 2
2
4 23 1 2 4
3
6 26 5 3 1 2 4
6
2 100000000002 1
2
Games in the second sample:
3 plays with 1. 3 wins. 1 goes to the end of the line.
3 plays with 2. 3 wins. He wins twice in a row. He becomes the winner.
题意: 打乒乓球,输入两个数 n,h,共有n 个人,h为要赢的场次数,下面一行为n个人的能量;先让前两个人开始打比赛,赢的人依次和下面的人打比赛,输的人在最后面排队,两个人输赢有两个人的能量定,能量高的人赢,输出第一个赢的h场的人;
思路:这是一道模拟题,数据范围可定义为 long long 首先要思考若轮完一轮了,还没有人赢h场,这时候一定是能量最高的那个人赢;还有在中间的人一定要和他的前一个人比(这一不要忘了)也和他后面的人比赛;记录每一个人第一轮能赢几场就行了,这道题主要就是不要只顾一边,前 后 都要顾,一定不要忘了和他前一个人相比;
代码:
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define Max 550long long a[Max],hou[Max];int main(){long long i,j,n,h;while(~scanf("%lld%lld",&n,&h)){long long Ma = -1;for(i=0;i<n;i++){scanf("%lld",&a[i]);Ma = max(Ma,a[i]);}for(i=0;i<n;i++){long long num ;if(i>0&&a[i]>a[i-1]) // 一定不要忘了和他前一个人相比; num = 1;else num = 0;for(j=i+1;j<n;j++){if(a[i]>a[j])num++;else break;}if(num>=h){printf("%lld\n",a[i]);break;}}if(i>=n)printf("%lld\n",Ma);}return 0;}
- codeforces~~~
- Codeforces
- codeforces
- Codeforces
- codeforces
- codeforces
- Codeforces
- Codeforces
- CodeForces
- CodeForces
- CodeForces
- CodeForces
- CodeForces
- Codeforces
- Codeforces
- Codeforces
- Codeforces
- Codeforces
- JuiceSSH
- Linux中使用yum安装jdk
- 数据库建表的主键问题
- String类、StringBuffer类和StringBuilder类的区别
- js 一些时间之间的转换
- CodeForces
- docker 出现 Error response from daemon
- Java集合类详解
- js中数组的一些使用
- TensorFlow学习 -2
- QPSK 调制与解调(Matlab仿真)
- C#实现对图片文件的压缩、裁剪操作实例
- 斐讯K2路由器刷固件实现校园网可使用
- mysql navicat编码保持一致不乱码