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.
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).
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.
4 2 3 1 6
2
4 2 3 1 7
4
1 2 3 2 6
13
1 1 2 1 1
0
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <vector> 6 #include <climits> 7 #include <algorithm> 8 #include <cmath> 9 #define LL long long 10 #define INF 0x3f3f3f 11 using namespace std; 12 double a,b,w,x,c; 13 int main(){ 14 while(~scanf("%lf %lf %lf %lf %lf",&a,&b,&w,&x,&c)){ 15 double ans = ceil((w*c-w*a-b)/(w-x)); 16 printf("%.0f\n",c<=a?0:ans); 17 } 18 return 0; 19 }
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xtu summer individual 6 B - Number Busters
原文：http://www.cnblogs.com/crackpotisback/p/3900281.html
