Where is Vasya?

Vasya stands in line with number of people p (including Vasya), but he doesn't know exactly which position he occupies. He can say that there are no less than b people standing in front of him and no more than apeople standing behind him. Find the number of different positions Vasya can occupy.

Input

As an input you have 3 numbers:

1. Total amount of people in the line;

2. Number of people standing in front of him

3. Number of people standing behind him

Examples

Line.WhereIsHe(3, 1, 1) // => 2 The possible positions are: 2 and 3
Line.WhereIsHe(5, 2, 3) // => 3 The possible positions are: 3, 4 and 5

The third parameter is not irrelavant and is the reason why (9,4,3) is 4 not 5
you have to satisfy both conditions
no less than bef people in front of him
and
no more than aft people behind him

as far as i can tell all the test cases are correct

9个人,前面的人不少于4个,后面的人不多于3个的话,可以占据6,7,8,9 是个位置

第五个位置,虽然前面是4个人,但是后面也是4个人。后面的人数超过3了,就不符合。

using System;

public class Line
{
public static int WhereIsHe(int p, int bef ,int aft)
{
// Your code is here...
int count=;
int a=;//people infront of him
int b=;//people behind him
for(int i=;i<=p;i++)
{
a=i-;
b=p-i;
if(a>=bef&&b<=aft)
{
count++;
}
}
return count;
}
}

使用Linq进行简化后:

using System;
using System.Linq;
public static int WhereIsHe(int p, int bef, int aft)
{
return Enumerable.Range(, p).Where(x => x - >= bef && p - x <= aft).Count();
}

其他人的解法

using System;

public class Line
{
public static int WhereIsHe(int p, int bef ,int aft)
{
return Math.Min(p-bef,aft+);
}
}

Where is Vasya?的更多相关文章

  1. Milliard Vasya's Function-Ural1353动态规划

    Time limit: 1.0 second Memory limit: 64 MB Vasya is the beginning mathematician. He decided to make ...

  2. CF460 A. Vasya and Socks

    A. Vasya and Socks time limit per test 1 second memory limit per test 256 megabytes input standard i ...

  3. 递推DP URAL 1353 Milliard Vasya's Function

    题目传送门 /* 题意:1~1e9的数字里,各个位数数字相加和为s的个数 递推DP:dp[i][j] 表示i位数字,当前数字和为j的个数 状态转移方程:dp[i][j] += dp[i-1][j-k] ...

  4. Codeforces Round #281 (Div. 2) D. Vasya and Chess 水

    D. Vasya and Chess time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

  5. Codeforces Round #281 (Div. 2) C. Vasya and Basketball 二分

    C. Vasya and Basketball time limit per test 2 seconds memory limit per test 256 megabytes input stan ...

  6. codeforces 676C C. Vasya and String(二分)

    题目链接: C. Vasya and String time limit per test 1 second memory limit per test 256 megabytes input sta ...

  7. Codeforces Round #324 (Div. 2) C. Marina and Vasya 贪心

    C. Marina and Vasya Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/584/pr ...

  8. Codeforces Round #322 (Div. 2) A. Vasya the Hipster 水题

    A. Vasya the Hipster Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/581/p ...

  9. Codeforces Codeforces Round #319 (Div. 2) C. Vasya and Petya's Game 数学

    C. Vasya and Petya's Game Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/ ...

随机推荐

  1. Projected Coordinate Systems

    Coordinate Systems Projected Coordinate Systems This is an archive of a previous version of the ArcG ...

  2. android系统网络信号强弱参数之ecsq指令参数说明

    ecsq指令参数说明: // For LTE (MT6592) // AT+ECSQ:<sig1>,<sig2>,<rssi_in_qdbm>,<rscp_i ...

  3. A3992学习记录

    ATmega64+A3992驱动步进电机 //ATmega 64a 电机驱动板程序//编译环境 AVR Studio 4.17/AVR GCC//系统外部时钟16M//作者:虞恺 //日期:2012. ...

  4. 四个基数任意次数组合相加得到一个数N,求所有可能组合

    #include <iostream> #include <vector> usingnamespace std; vector<int> vec; constin ...

  5. 九,WPF资源

    WPF资源的优点 WPF资源系统是一种保管一系列有用对象的简单方法,从而可以更容易地重用这些对象,它主要有以下优点: 高效,通过资源可以定义一个对象,并在标记中的多个地方重用,这会使代码变的更加精简, ...

  6. 1046 Shortest Distance (20)

    #include<stdio.h> int main() { int n,m,a,b,tem,pre,p; int i,j; ]; while(scanf("%d",& ...

  7. Qt 内存管理机制(转)

      许转载http://devbean.blog.51cto.com/448512/526734 强类型语言在创建对象时总会显式或隐式地包含对象的类型信息.也就是说,强类型语言在分配对象内存空间时,总 ...

  8. The method of type must override a superclass method

    导入android项目时,报The method of type must override asuperclass method 一堆错误, 解决方法: 将编译的jdk与使用的jdk版本一致即可.

  9. Nginx配置文件变量大全

    $args # 这个变量等于请求行中的参数. $binary_remote_addr # 远程地址的二进制表示 $body_bytes_sent # 已发送的消息体字节数 $content_lengt ...

  10. 【学习总结】UIGestureRecognizer(手势识别器)

    基本知识点 : -> IOS 3.2之后 , 苹果推出了手势识别功能 ( Gesture Recognizer ) 在触摸事件处理方面 , 简化开发难度. -> UIGesture Rec ...