2013 Asia Chengdu Regional Contest J

Just Random

Time Limit: 2000/1000 MS (Java/Others)   

Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

Coach Pang and Uncle Yang both love numbers. Every morning they play a game with number together. In each game the following will be done: 1. Coach Pang randomly choose a integer x in [a, b] with equal probability. 2. Uncle Yang randomly choose a integer y in [c, d] with equal probability. 3. If (x + y) mod p = m, they will go out and have a nice day together. 4. Otherwise, they will do homework that day. For given a, b, c, d, p and m, Coach Pang wants to know the probability that they will go out.

 

Input

The first line of the input contains an integer T denoting the number of test cases. For each test case, there is one line containing six integers a, b, c, d, p and m(0 <= a <= b <= 10⁹, 0 <=c <= d <= 10⁹, 0 <= m < p <= 10⁹).

 

Output

For each test case output a single line "Case #x: y". x is the case number and y is a fraction with numerator and denominator separated by a slash ('/') as the probability that they will go out. The fraction should be presented in the simplest form (with the smallest denominator), but always with a denominator (even if it is the unit).

 

Sample Input

4

0 5 0 5 3 0

0 999999 0 999999 1000000 0

0 3 0 3 8 7

3 3 4 4 7 0

 

Sample Output

Case #1: 1/3

Case #2: 1/1000000

Case #3: 0/1

Case #4: 1/1

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
#define ll __int64
ll gcd(ll n, ll m){ return m ? gcd(m, n%m) : n; }
ll n, m, a, b, c, d, p,t1, t2, cnt1, cnt2, add, sub,s;
int main(){
    int t, cas = 1;
    scanf("%d", &t);
    while (t--){
        s = 0;
        scanf("%I64d%I64d%I64d%I64d%I64d%I64d", &a, &b, &c, &d, &p, &m);
        if (b + c <= a + d)swap(a, c), swap(b, d);
        t1 = (a + c) % p;
        add = (m - t1 + p) % p;
        cnt1 = (a+c + add - m) / p;
        t2 = (b + c-1) % p;
        sub = (t2 - m + p) % p;
        cnt2 = (b + c-1 - sub - m) / p;
        s += (cnt2 - cnt1 + 1)*(add + 1) + (cnt2 - cnt1 + 1)*(cnt2 - cnt1) / 2 * p;
        t1 = (b + c) % p;
        add = (m - t1 + p) % p;
        cnt1 = (b + c + add - m) / p;
        t2 = (a + d) % p;
        sub = (t2 - m + p) % p;
        cnt2 = (a + d - sub - m) / p;
        s += (cnt2 - cnt1 + 1)*(b - a + 1);
        t1 = (a + d + 1) % p;
        add = (m - t1 + p) % p;
        cnt1 = (a + d + 1+add-m) / p;
        t2 = (b + d) % p;
        sub = (t2 - m + p) % p;
        cnt2 = (b + d - sub - m) / p;
        s += (cnt2 - cnt1 + 1)*(sub + 1) + (cnt2 - cnt1 + 1)*(cnt2 - cnt1) / 2 * p;
        ll ss = (b - a + 1)*(d - c + 1);
        printf("Case #%d: ", cas++);
        printf("%I64d/%I64d\n", s / gcd(s, ss), ss / gcd(s, ss));
    }
    return 0;
}