Codeforces Round #298 (Div. 2) B. Covered Path 物理题/暴力枚举

B. Covered Path

Time Limit: 1 Sec  Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/534/problem/B

Description

The on-board computer on Polycarp's car measured that the car speed at the beginning of some section of the path equals v₁ meters per second, and in the end it is v₂ meters per second. We know that this section of the route took exactly t seconds to pass.

Assuming that at each of the seconds the speed is constant, and between seconds the speed can change at most by d meters per second in absolute value (i.e., the difference in the speed of any two adjacent seconds does not exceed d in absolute value), find the maximum possible length of the path section in meters.

Input

The first line contains two integers v₁ and v₂ (1 ≤ v₁, v₂ ≤ 100) — the speeds in meters per second at the beginning of the segment and at the end of the segment, respectively.

The second line contains two integers t (2 ≤ t ≤ 100) — the time when the car moves along the segment in seconds, d (0 ≤ d ≤ 10) — the maximum value of the speed change between adjacent seconds.

It is guaranteed that there is a way to complete the segment so that:

• the speed in the first second equals v₁,
• the speed in the last second equals v₂,
• the absolute value of difference of speeds between any two adjacent seconds doesn't exceed d.

Output

Print the maximum possible length of the path segment in meters.

Sample Input

input1

5 6
4 2

input2

10 10
10 0

 

Sample Output

26

 

100

HINT

题意

啊，给你初速度和末速度，给你运动时间，和最大加速度

当然，每一秒的速度都是整数，然后问你在这种情况下，路程最大是多少

题解：

  啊，暴力枚举每秒钟加多少就好啦，肯定是越早加速越好啦

代码：

 

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
//const int inf=0x7fffffff;   //无限大
const int inf=0x3f3f3f3f;
/*
inline ll read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int buf[10];
inline void write(int i) {
  int p = 0;if(i == 0) p++;
  else while(i) {buf[p++] = i % 10;i /= 10;}
  for(int j = p-1; j >=0; j--) putchar('0' + buf[j]);
  printf("\n");
}
*/
//**************************************************************************************

int main()
{
    int v1,v2;
    cin>>v1>>v2;
    int d,t;
    cin>>t>>d;
    int ans=v1;
    for(int i=2;i<=t;i++)
    {
        for(int j=d;j>=-d;j--)
        {
            if(v1+j-(t-i)*d<=v2)
            {
                v1=v1+j;
                break;
            }
        }
        ans+=v1;
        //cout<<v1<<endl;
    }
    cout<<ans<<endl;
}