CodeForces 534B Covered Path

最新推荐文章于 2019-02-23 19:16:38 发布

Coder_Jeremy Wu

最新推荐文章于 2019-02-23 19:16:38 发布

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分类专栏： Codeforces 文章标签： codeforces

本文链接：https://blog.csdn.net/Coder_Jeremy/article/details/52024772

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本文介绍了一个有趣的问题：如何计算在给定的速度变化限制下，车辆从初始速度到最终速度能够行驶的最大距离。通过逐步调整速度并确保每秒内的速度变化不超过设定值，文章提供了一种算法来解决这一问题。

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Covered Path

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

Input

5 6
4 2

Output

Input

10 10
10 0

Output

100

Hint

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 = 26 meters.

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.

这题非常有意思，题意是这样的：
给定初始速度v1和终止速度v2，时间t，速度变化量d，求在时间t内运动的最大路程是多少。
题意很好理解，但是算法就很有意思了，自己没有想出来怎么算，在网上搜了一下，大致看懂了，然后自己又改写了一下。
思路：
因为在单位时间内速度变化量是有上限的，同时还要考虑路程最大，因此要尽量保证速度变化量最大。
如果保持速度始终是最大增量的话，在t时间内如果v1一直在增加，v2也在反向增加，此时都以最大速度变化量在改变速度，但实际情况是以该时间下速度较小的那一个为基准。
例如
v1=5 v2=6
t=4 d=2
v1变化， 5>>7>>9>>11
v2变化，12<<10<< 8<< 6
对比可得出每一秒下的最大速度分别为5 7 8 6
这应该是最好的算法！

AC

#include <bits/stdc++.h>
using namespace std;
const int maxn=105;

int main()
{
    int v1,v2,t,d;
    scanf("%d%d%d%d",&v1,&v2,&t,&d);
    int a[maxn],b[maxn];
    a[1]=v1;
    b[t]=v2;
    for(int i=2;i<=t;i++)
        a[i]=a[i-1]+d;
    for(int i=t-1;i>0;i--)
        b[i]=b[i+1]+d;
    int sum = 0;
        for(int i = 1; i <= t; i++)
            sum += min(a[i], b[i]);
        printf("%d", sum);//取速度较小的为基准
    return 0;
}