CF 382B Number Busters

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.

Example
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

【题意】

现在有五个数据a,b,w,x,c。每一分钟都会产生下面的效果：

c--;if(b>=x)    b-=x;else{    a--;    b=w-(x-b);}

问你经过多少秒之后会出现a>=c的情况。

【分析】
数学问题吧。推导过程如下：
首先把上面的化简一下：

c--;if(b>=x)    b-=x;else {    a--;    b+=w-x;}

然后我们发现c下降的永远比a快，所以上面的过程等价于：

if(b>=x){    c--;    b-=x;}else    b+=w-x;

这样总共需要的变换次数就显而易见为c-a。然后我们发现c的减少只出现在b>=x的时候，也就是b减少的时候，b什么时候增加呢？显然是b < x的时候。所以容易发现整个过程其实就是一个b增加然后减少的过程。就是询问需要增减的步数。

所以代码就显而易见了。
【代码】

#include <cstdio>#include <cstdlib>#include <cstring>#include <queue>using namespace std;long long my_floor(long long a,long long b) {    if(a%b) return a/b+1;    return a/b;}int main() {//    freopen("in.txt","r",stdin);    long long a,b,w,x,c;    while(scanf("%lld%lld%lld%lld%lld",&a,&b,&w,&x,&c)!=EOF) {        if(c<=a) {            printf("0\n");            continue;        }        long long need = c-a;        long long t1 = my_floor(need*x - b,w - x);        long long t2 = need;        printf("%lld\n",t1+t2);    }    return 0;}