HDU 4790 Just Random （2013成都J题）

Just Random

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 87    Accepted Submission(s): 34

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

 

Source

2013 Asia Chengdu Regional Contest

 

这题就是要找在[a,b]  [c,d] 之间，和模p等于m的对数。

把[a,b] [c,d]所有可能组合的和写成下列形式。

a+c  a+c+1  a+c+2   ..................a+d

a+c+1  a+c+2  a+c+3 ........a+d  a+d+1

a+c+2  a+c+3         a+d   a+d+1   a+d+2

....................

...................

b+c   b+c+1   ...............................................b+d;

上面大致形成一个斜的矩阵。

使用b+c  和 a+d两条竖线，就可以分成三部分。前后两部分个数是等差数列，中间个数是相等的。

只需要讨论下b+c 和 a+d的大小。  然后找到%p==m 的位置，求和就可以搞定了。

 /* ***********************************************
 Author        :kuangbin
 Created Time  :2013-11-16 13:20:40
 File Name     :E:\2013ACM\专题强化训练\区域赛\2013成都\1010.cpp
 ************************************************ */
 
 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;
 
 long long gcd(long long a,long long b)
 {
     if(b == )return a;
     return gcd(b,a%b);
 }
 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     long long a,b,c,d,p,m;
     int T;
     int iCase = ;
     scanf("%d",&T);
     while(T--)
     {
         iCase++;
         scanf("%I64d%I64d%I64d%I64d%I64d%I64d",&a,&b,&c,&d,&p,&m);
         long long ans = ;
         if(b+c <= a+d)
         {
             long long t1 = (a+c)%p;
             long long add = (m - t1 + p)%p;
             long long cnt1 = (a+c + add-m)/p;
             //cout<<t1<<" "<<add<<endl;
             long long t2 = (b+c-)%p;
             long long sub = (t2 - m + p)%p;
             long long cnt2 = (b+c--sub-m)/p;
             //cout<<t2<<" "<<sub<<endl;
             ans += (cnt2 - cnt1 + )*(+add) + (cnt2 - cnt1 + )*(cnt2 - cnt1)/ * p;
             //printf("%I64d %I64d  %I64d\n",cnt1,cnt2,ans);
             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;
             ans += (cnt2 - cnt1 + )*(b-a+);
             t1 = (a+d+)%p;
             add = (m - t1 + p)%p;
             cnt1 = (a+d++add-m)/p;
             t2 = (b+d)%p;
             sub = (t2 - m + p)%p;
             cnt2 = (b+d-sub-m)/p;
             ans += (cnt2 - cnt1 + )*(+sub) + (cnt2 - cnt1 + )*(cnt2 - cnt1)/*p;
         }
         else
         {
             long long t1 = (a+c)%p;
             long long add = (m - t1 + p)%p;
             long long cnt1 = (a+c + add-m)/p;
             long long t2 = (a+d-)%p;
             long long sub = (t2 - m + p)%p;
             long long cnt2 = (a+d--sub-m)/p;
             ans += (cnt2 - cnt1 + )*(+add) + (cnt2 - cnt1 + )*(cnt2 - cnt1)/ * p;
             t1 = (a+d)%p;
             add = (m - t1 + p)%p;
             cnt1 = (a+d+add-m)/p;
             t2 = (b+ c)%p;
             sub = (t2 - m + p)%p;
             cnt2 = (b+c-sub-m)/p;
             ans += (cnt2 - cnt1 + )*(d-c+);
             t1 = (b+c+)%p;
             add = (m - t1 + p)%p;
             cnt1 = (b+c++add-m)/p;
             t2 = (b+d)%p;
             sub = (t2 - m + p)%p;
             cnt2 = (b+d - sub-m)/p;
             ans += (cnt2 - cnt1 + )*(+sub) + (cnt2 - cnt1 + )*(cnt2 - cnt1)/*p;
         }
         long long tot = (b-a+)*(d-c+);
         long long GCD = gcd(ans,tot);
         ans /= GCD;
         tot /= GCD;
         printf("Case #%d: %I64d/%I64d\n",iCase,ans,tot);
     }
     return ;
 }