cf 534B. Covered Path

B. Covered Path

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

input

5 6
4 2

output

26

input

10 10
10 0

output

100

Note

In the first sample the sequence of speeds of Polycarpus' car can look as follows: 5, 7, 8, 6. Thus, the total path is 5 + 7 + 8 + 6 = 26meters.

In the second sample, as d = 0, the car covers the whole segment at constant speed v = 10. In t = 10 seconds it covers the distance of 100 meters.

题意是说一辆车，每秒内的速度恒定...第I秒到第I+1秒的速度变化不超过D。初始速度为V1，末速度为V2，经过时间t，问最远能走多远。

策略就是尽可能加速...加到某个时间，如果在这个时间不开始减速就回不到V2了。

从后往前预处理下每秒钟能达到的最大速度（如果超过这个速度，将不能减回到V2）

#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>

using namespace std;
const int N=1E2+5;
const int inf=8E9;
int a[N],m[N];
int v1,v2,t,d,tmp,p,ans,n;


int main()
{
    cin>>v1>>v2>>t>>d;
    p = inf;
    memset(a,0,sizeof(a));
    a[1]= v1;
    tmp = v2;
    m[t]= v2;
    for ( int i = t-1 ; i >= 1; i--)
    {
        tmp = tmp+d;
        m[i] = tmp;
    }
    for ( int i =2 ;i <= t; i++ )
    {
        if (a[i-1]+d<=m[i])
        {
            a[i] = a[i-1] + d;
        }
        else
        {
            a[i] = m[i];
            p = i;
            break;
        }
    }
    ans = 0;
    for ( int i = p ; i <= t; i++ )
        a[i] = m[i];
    for ( int i = 1; i <= t ; i++ )
    {
        if (i<p)
            ans = ans + a[i];
        else ans = ans + m[i];
    }
   // for ( int i = 1; i <= t ; i++)
   //     cout<<"a[i]:"<<a[i]<<endl;
    cout<<ans<<endl;

    return 0;
}