Xtu summer individual 6 B-number busters

Number busterstime limit: 1000 msmemory limit: 262144 kbthis problem will be judged on codeforces. Original ID: 382b
64-bit integer Io format: % i64d Java class name: (any) Arthur and Alexander are number Busters. Today they 've got a competition.

 

Arthur took a group of four IntegersA, Bytes,B, Bytes,W, Bytes,X(0 bytes ≤ bytesBLatency <latencyW, When 0 then <thenXLatency <latencyW) And Alexander took integerBytes. 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:CSignature = SignatureCExample-Example 1. Arthur is a sophisticated guy. Each second Arthur performs a complex operation, described as follows: ifBLimit ≥ limitX, Perform the assignmentBSignature = SignatureBAccept-Encoding-X, IfBLatency <latencyX, Then perform two consecutive assignmentsASignature = SignatureAAccept-limit 1;BSignature = SignatureWAudio-extract -(XAccept-Encoding-B).

You 've got numbersA, Bytes,B, Bytes,W, Bytes,X, Bytes,C. Determine when Alexander gets ahead of Arthur if both guys start making the operations at the same time. Assume that Alexander got ahead of Arthur ifCLimit ≤ limitA.

Input

The first line contains IntegersA, Bytes,B, Bytes,W, Bytes,X, Bytes,C(1 digit ≤ DigitALimit ≤ limit 2 · 109, limit 1 limit ≤ limitWLimit ≤ limit 1000, limit 0 limit ≤ limitBLatency <latencyW, When 0 then <thenXLatency <latencyW, Memory 1 ≤ memoryCLimit ≤ limit 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 inputinput

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

Sourcecodeforces round #224 (div. 2) Problem Solving: Assuming c '= c-t after T times; a' = A-N; B' = B-Tx + NW; after the c '<= A' is solved, there is: (WC-wa-B + B')/(W-x) <= t because t needs to be rounded up, and the minimum value is, therefore, when B 'is equal to 0, T has a minimum lower bound ....

 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 long10 #define INF 0x3f3f3f11 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 }

View code