CF 224DIV2 B Number Busters

B. Number Busters

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

2

input

4 2 3 1 7

output

4

input

1 2 3 2 6

output

13

input

1 1 2 1 1

output

0

这题直接循环肯定会TLE,所以我们需要hash记录b，一次来找出循环节

比赛的时候，这题我一直WA,可惜就是找不出错误，最后才发现原来是最大循环超int型了, 比赛之后一下就找出了，，，哭瞎了都

代码：

#include<stdio.h>#include<math.h>#include<string.h>#include<algorithm>#include<limits.h>using namespace std;int hash[1111];int hash2[1111];int main(){int a,b,w,x,c;while(scanf("%d %d %d %d %d",&a,&b,&w,&x,&c)!=EOF){memset(hash,-1,sizeof(hash));int d= w- x;__int64 ans= 0;int sum= 0; // b出现的不同数的个数 int flag= 0;while(c> a){int max= b/x;//printf("%d\n",max);if(c- max<= a){ans+= c- a;c= a;break;}else{c-= max;b-= max * x;ans+= max;if(flag==0){if(hash[b]!=-1){__int64 hehe= ans- hash[b];__int64 me= (c - a )/ (hash2[b]- c);//printf("%d %d\n",me,hehe);ans+= me* hehe; // 此处会超 int!!! //printf("%I64d\n",ans);c-= me *(hash2[b]- c); flag= 1;}else{hash[b]= ans;hash2[b]= c;}}if(c<= a)break;int k= (x- b)/ d;b= b+ k*d;ans+= k;if(b< x){b+= d;ans++;}}}printf("%I64d\n",ans);}return 0;}