CF Covered Path (贪心)

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

哈哈，终于可以完全理解的A出一道贪心题了。假设每个点的速度是V,剩余的时间是T_S，加速量是X，那么需要满足 V + X <= V_2 + T_S * d,以此来更新每个点的速度就行。

 #include <iostream>
 #include <cstdio>
 using    namespace    std;

 int    main(void)
 {
     int    v_1,v_2,t,d;
     int    sum = ;

     cin >> v_1 >> v_2;
     cin >> t >> d;
     sum += v_1;

     int    v = v_1;
     for(int i = ;i <= t;i ++)
         for(int j = d;j >= -d;j --)
             if(v + j <= v_2 + (t - i) * d)
             {
                 v += j;
                 sum += v;
                 break;
             }
     cout << sum << endl;

     return    ;
 }