Just Random HDU - 4790 思维题（打表找规律）分段求解

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

　　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("******\n")
 #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 + ;
 const int maxn = 1e5 + ;
 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 ;
     else {
         int temp = cnt - ( a + c ) + ;
         int num = ( R - cnt ) / p + ;
         LL sum = 1LL * temp * num + 1LL * num * ( num -  ) * p / ;
         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  ;
     else {
         LL temp = L - ( a + c ) + ;
         LL num = ( R - cnt ) / p + ;
         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  ;
     else {
         LL temp =  ( b + d ) - cnt  + ;
         LL num = ( R - cnt ) / p + ;
         LL sum = 1LL * temp * num - 1LL * num * ( num -  ) *  p / ;
         return sum;
     }
 }
 int main() {
     sf ( _ );
     int cas = ;
     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 = ;
         ans += solveL ( L, l -  );
         ans += solveM ( l, r );
         ans += solveR ( r + , R );
         LL sum = ( b - a +  ) * ( d - c +  );
         LL g = gcd ( ans, sum );
         printf ( "Case #%d: %lld/%lld\n", cas++, ans / g, sum / g );
     }
     return ;
 }