Just Random HDU

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 <= 10 ⁹, 0 <=c <= d <= 10 ⁹, 0 <= m < p <= 10 ⁹). 
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

给出 a<=x<=b c<=y<=d 求满足 （x+y） % p ==m 的概率
这题需要找规律 然后根据规律求解 
每一个数出现的次数为 上升的等差数列 相同的值 下降的等差数列 
以下看规律

0 5 0 5 3 0
0 1 1 2 2 2 3 3 3 3 4 4 4 4 4 5 5 5 5 5 5 6 6 6 6 6 7 7 7 7 8 8 8 9 9 10
3 7 2 5 4 6
5 6 6 7 7 7 8 8 8 8 9 9 9 9 10 10 10 11 11 12
1 3 2 6 2 5
3 4 4 5 5 5 6 6 6 7 7 7 8 8 9

　　然后根据等差数列求和 求解

#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <set>
#include <iostream>
#include <map>
#include <stack>
#include <string>
#include <vector>
#define  pi acos(-1.0)
#define  eps 1e-6
#define  fi first
#define  se second
#define  lson l,m,rt<<1
#define  rson m+1,r,rt<<1|1
#define  rtl   rt<<1
#define  rtr   rt<<1|1
#define  bug         printf("******
")
#define  mem(a,b)    memset(a,b,sizeof(a))
#define  name2str(x) #x
#define  fuck(x)     cout<<#x" = "<<x<<endl
#define  f(a)        a*a
#define  sf(n)       scanf("%d", &n)
#define  sff(a,b)    scanf("%d %d", &a, &b)
#define  sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define  sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
#define  pf          printf
#define  FRE(i,a,b)  for(i = a; i <= b; i++)
#define  FREE(i,a,b) for(i = a; i >= b; i--)
#define  FRL(i,a,b)  for(i = a; i < b; i++)
#define  FRLL(i,a,b) for(i = a; i > b; i--)
#define  FIN         freopen("in.txt","r",stdin)
#define  gcd(a,b)    __gcd(a,b)
#define  lowbit(x)   x&-x
using namespace std;
typedef long long  LL;
typedef unsigned long long ULL;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
int _;
LL  a, b, c, d, m, p;
LL solveL ( LL L, LL R ) {
    LL cnt = L % p;
    if ( cnt > m )  cnt = L + p - cnt + m;
    else cnt = L + m - cnt ;
    if ( cnt > R ) return 0;
    else {
        int temp = cnt - ( a + c ) + 1;
        int num = ( R - cnt ) / p + 1;
        LL sum = 1LL * temp * num + 1LL * num * ( num - 1 ) * p / 2;
        return sum;
    }
}
LL solveM ( LL L, LL R ) {
    LL cnt = L % p;
    if ( cnt > m )  cnt = L + p - cnt + m;
    else cnt = L + m - cnt ;
    if ( cnt > R ) return 0 ;
    else {
        LL temp = L - ( a + c ) + 1;
        LL num = ( R - cnt ) / p + 1;
        LL sum = 1LL * temp * num;
        return sum;
    }
}
LL solveR ( LL L, LL R ) {
    LL cnt = L % p;
    if ( cnt > m )  cnt = L + p - cnt + m;
    else cnt = L + m - cnt ;
    if ( cnt > R ) return 0 ;
    else {
        LL temp =  ( b + d ) - cnt  + 1;
        LL num = ( R - cnt ) / p + 1;
        LL sum = 1LL * temp * num - 1LL * num * ( num - 1 ) *  p / 2;
        return sum;
    }
}
int main() {
    sf ( _ );
    int cas = 1;
    while ( _-- ) {
        scanf ( "%lld%lld%lld%lld%lld%lld", &a, &b, &c, &d, &p, &m );
        LL L = a + c, R = b + d, l = min ( b + c, a + d ), r = max ( b + c, a + d ), ans = 0;
        ans += solveL ( L, l - 1 );
        ans += solveM ( l, r );
        ans += solveR ( r + 1, R );
        LL sum = ( b - a + 1 ) * ( d - c + 1 );
        LL g = gcd ( ans, sum );
        printf ( "Case #%d: %lld/%lld
", cas++, ans / g, sum / g );
    }
    return 0;
}