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：车在开始那一秒和结束那一秒的速度为V1,V2 ，给出总共行驶的时间，而且每秒的速度增加与减少最多为d，要求出最大的行驶距离
思路：我们假设最高速度为第i秒时，那么从1~i是一个以v1为起始项的的递增序列，而从t~i则是以v2为起始项的递增序列，那么枚举i的位置求出来的值必然是最大的
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <algorithm>
using namespace std;
#define ls 2*i
#define rs 2*i+1
#define up(i,x,y) for(i=x;i<=y;i++)
#define down(i,x,y) for(i=x;i>=y;i--)
#define mem(a,x) memset(a,x,sizeof(a))
#define w(a) while(a)
#define LL long long
const double pi = acos(-1.0);
#define Len 63
#define mod 19999997
const int INF = 0x3f3f3f3f;
int v1,v2,t,d,sum;
int main()
{
int i,j,k;
w(~scanf("%d%d%d%d",&v1,&v2,&t,&d))
{
sum = 0;
up(i,0,t-1)
sum+=min(v1+i*d,v2+(t-i-1)*d);
printf("%d\n",sum);
}
return 0;
}