CF contest/382/B. Number Busters

Arthur与Alexander的数字竞赛

最新推荐文章于 2022-06-20 22:53:00 发布

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本文探讨了Arthur和Alexander通过不同策略减少整数的过程。Alexander每秒递减1，而Arthur的操作更为复杂。文章推导出Arthur保持领先Alexander所需的最短时间，并提供了完整的算法实现。

题解:

设t 分钟后：

c'=c-t;

a'=a- (x*t-b)/w;

c'<=a';

整理出来 t>= (a*w-cw+b)/(x-w);

注意题目条件 x<w;如果没有则要考虑x==w ,x>w的情况；

关于这个 a'=a- (x*t-b)/w; 一开始以为大家一看公式就明白，所以没有解释，抱歉~

t分钟的后， Arthur 一共减去了x*t，对吧？ x*t减去最初的b，剩下的是需要改变a值来继续游戏的。所以看x*t-b 可以分成多少个w ,(w是a-1 转化来的) 因此a的改变量就是(x*t-b)/w;

不知道是否解释清楚，还有疑问，请留言^.^。

题目： Arthur took a group of four integers a, b, w, x (0 ≤ b < w, 0 < x < w) and Alexander took integer с. Arthur and Alexander use distinct approaches to number bustings. Alexander is just a regular guy. Each second, he subtracts one from his number. In other words, he performs the assignment: c = c - 1. Arthur is a sophisticated guy. Each second Arthur performs a complex operation, described as follows: if b ≥ x, perform the assignment b = b - x, if b < x, then perform two consecutive assignments a = a - 1; b = w - (x - b).

You've got numbers a, b, w, x, c. Determine when Alexander gets ahead of Arthur if both guys start performing the operations at the same time. Assume that Alexander got ahead of Arthur if c ≤ a.

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
long long a,b,w,x,c;//  比赛时候一激动竟然定义成int;脑残！
long long ans;
void deal()
{
        if(c<=a) {ans=0;return;}
        long long  pt=(1ll*a*w-1ll*w*c+b)/(x-w);
        ans=pt;
        if(abs(1ll*pt*(x-w))<abs(1ll*a*w-1ll*w*c+b))ans++;
        return;
}
int main()
{
// freopen("in.txt","r",stdin);

 while(~scanf("%I64d%I64d%I64d%I64d%I64d",&a,&b,&w,&x,&c))
 {
   deal();
   printf("%I64d\n",ans);
 }

 return 0;
}