codeforces 1066a Vova and Train

最新推荐文章于 2019-03-07 15:58:59 发布

原创最新推荐文章于 2019-03-07 15:58:59 发布 · 552 阅读

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#数学 #思维

本文探讨了一道关于火车路径上可见灯塔数量的算法题。火车从起点出发，途中遇到每隔v个单位距离设置的灯塔，但由于固定停靠的火车遮挡，部分灯塔不可见。文章通过数学分析解决了这一问题，并提供了有效的代码实现。

Vova plans to go to the conference by train. Initially, the train is at the point 1 and the destination point of the path is the point L. The speed of the train is 1 length unit per minute (i.e. at the first minute the train is at the point 1, at the second minute — at the point 2 and so on).

There are lanterns on the path. They are placed at the points with coordinates divisible by v (i.e. the first lantern is at the point v, the second is at the point 2v and so on).

There is also exactly one standing train which occupies all the points from l to r inclusive.

Vova can see the lantern at the point p if p
p is divisible by v and there is no standing train at this position (p∉[l;r]). Thus, if the point with the lantern is one of the points covered by the standing train, Vova can’t see this lantern.

Your problem is to say the number of lanterns Vova will see during the path. Vova plans to go to t
t different conferences, so you should answer t independent queries.

题意很好懂，分析下来就是一个数学问题。
一开始用的模拟，超时了，嘤嘤嘤。在分析一下就是数学问题，好好分析一下。
代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;

ll dis,v,l,r;

int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%I64d%I64d%I64d%I64d",&dis,&v,&l,&r);
		ll sum=dis/v;
		ll ans=0;
		ll mod=l%v;
		if(mod==0)
		{
			printf("%I64d\n",sum-((r-l)/v+1));
		}
		else
		{
			printf("%I64d\n",sum-(r-l+mod)/v);
		}
	}
}

努力加油a啊，(^o)/~