cf——B. Number Busters

2024-06-09 04:58

文章标签 number cf busters

本文主要是介绍cf——B. Number Busters，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

Arthur and Alexander are number busters. Today they've got a competition.

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.

Input

The first line contains integers a, b, w, x, c (1 ≤ a ≤ 2·109, 1 ≤ w ≤ 1000, 0 ≤ b < w, 0 < x < w, 1 ≤ c ≤ 2·109).

Output

Print a single integer — the minimum time in seconds Alexander needs to get ahead of Arthur. You can prove that the described situation always occurs within the problem's limits.

Sample test(s)

input

4 2 3 1 6

output

input

4 2 3 1 7

output

input

1 2 3 2 6

output

input

1 1 2 1 1

output 

第一次做的时候觉得这个题没自己想的那么简单，但还是试了一下，果然超时了 

#include<iostream>
#include<cstring>
#include<cstdio>
#include <algorithm>
using namespace std;
int main()
{int a,b,w,x,c;scanf("%d%d%d%d%d",&a,&b,&w,&x,&c);int  t=0;while(c>a){if(b>=x)b=b-x;else if(b<x){a--;b=w-(x-b);}t++;c=c-1;}printf("%d\n",t);return 0;
}

后来，看了一下别人的公式法，比较给力

意：求经过多少时间c<=a，c和a都有变换的方式。

思路：推个公式。。c每秒都减1，a在b<x的时候减1,并且b = b + w - x,在b>=x的时候不变化，并且b = b - x..也就是a和c只有在b = b - x的时候差距才会减少1,所以b一共减了(c-a)次x, 设加了k次w - x.那么得到b最后的值为b - x * (c - a) + k * (w - x).然后b最后的值+x 必然不小于x(因为b >= x 才是执行b = b - x)..所以得到公式b - x * (c - a) + k *(w - x) + x >= x ==> k = (ceil)((b - x * (c - a))/(w - x)).

#include <iostream>
#include <cstdio>
#include<cmath>
#include <cstring>
using namespace std;long long a,b,w,x,c;
long long fun()
{if(c<=a)return 0;long long ans=(long long )ceil((x*(c-a)-b)*1.0/(w-x))+(c-a);return ans;
}
int main()
{scanf("%d%d%d%d%d",&a,&b,&w,&x,&c);cout<<fun()<<endl;return 0;
}

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