B. Number Busters

http://codeforces.com/contest/382/problem/B

题解:

   设t 分钟时：

    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；

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.

code:

#include<stdio.h>  
#include<string.h>  
#include<algorithm>  
#include "cmath"
using namespace std;  
long long a,b,w,x,c;
long long ans;  
void deal()  
{  
        if(c<=a) {ans=0;return;}  
        long long pt=(ceil)((double)(a*w-w*c+b)/(double)(x-w));  
       
        ans=pt;  
        
        //if(abs(pt*(x-w))<abs(a*w-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;  
}