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
 

 

 

 

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。

 

 1 /* ***********************************************
 2 Author        :kuangbin
 3 Created Time  :2013-11-16 13:20:40
 4 File Name     :E:\2013ACM\专题强化训练\区域赛\2013成都\1010.cpp
 5 ************************************************ */
 6 
 7 #include <stdio.h>
 8 #include <string.h>
 9 #include <iostream>
10 #include <algorithm>
11 #include <vector>
12 #include <queue>
13 #include <set>
14 #include <map>
15 #include <string>
16 #include <math.h>
17 #include <stdlib.h>
18 #include <time.h>
19 using namespace std;
20 
21 long long gcd(long long a,long long b)
22 {
23     if(b == 0)return a;
24     return gcd(b,a%b);
25 }
26 int main()
27 {
28     //freopen("in.txt","r",stdin);
29     //freopen("out.txt","w",stdout);
30     long long a,b,c,d,p,m;
31     int T;
32     int iCase = 0;
33     scanf("%d",&T);
34     while(T--)
35     {
36         iCase++;
37         scanf("%I64d%I64d%I64d%I64d%I64d%I64d",&a,&b,&c,&d,&p,&m);
38         long long ans = 0;
39         if(b+c <= a+d)
40         {
41             long long t1 = (a+c)%p;
42             long long add = (m - t1 + p)%p;
43             long long cnt1 = (a+c + add-m)/p;
44             //cout<<t1<<" "<<add<<endl;
45             long long t2 = (b+c-1)%p;
46             long long sub = (t2 - m + p)%p;
47             long long cnt2 = (b+c-1-sub-m)/p;
48             //cout<<t2<<" "<<sub<<endl;
49             ans += (cnt2 - cnt1 + 1)*(1+add) + (cnt2 - cnt1 + 1)*(cnt2 - cnt1)/2 * p;
50             //printf("%I64d %I64d  %I64d\n",cnt1,cnt2,ans);
51             t1 = (b+c)%p;
52             add = (m - t1 + p)%p;
53             cnt1 = (b+c+add-m)/p;
54             t2 = (a+d)%p;
55             sub = (t2 - m + p)%p;
56             cnt2 = (a+d-sub-m)/p;
57             ans += (cnt2 - cnt1 + 1)*(b-a+1);
58             t1 = (a+d+1)%p;
59             add = (m - t1 + p)%p;
60             cnt1 = (a+d+1+add-m)/p;
61             t2 = (b+d)%p;
62             sub = (t2 - m + p)%p;
63             cnt2 = (b+d-sub-m)/p;
64             ans += (cnt2 - cnt1 + 1)*(1+sub) + (cnt2 - cnt1 + 1)*(cnt2 - cnt1)/2*p;
65         }
66         else
67         {
68             long long t1 = (a+c)%p;
69             long long add = (m - t1 + p)%p;
70             long long cnt1 = (a+c + add-m)/p;
71             long long t2 = (a+d-1)%p;
72             long long sub = (t2 - m + p)%p;
73             long long cnt2 = (a+d-1-sub-m)/p;
74             ans += (cnt2 - cnt1 + 1)*(1+add) + (cnt2 - cnt1 + 1)*(cnt2 - cnt1)/2 * p;
75             t1 = (a+d)%p;
76             add = (m - t1 + p)%p;
77             cnt1 = (a+d+add-m)/p;
78             t2 = (b+ c)%p;
79             sub = (t2 - m + p)%p;
80             cnt2 = (b+c-sub-m)/p;
81             ans += (cnt2 - cnt1 + 1)*(d-c+1);
82             t1 = (b+c+1)%p;
83             add = (m - t1 + p)%p;
84             cnt1 = (b+c+1+add-m)/p;
85             t2 = (b+d)%p;
86             sub = (t2 - m + p)%p;
87             cnt2 = (b+d - sub-m)/p;
88             ans += (cnt2 - cnt1 + 1)*(1+sub) + (cnt2 - cnt1 + 1)*(cnt2 - cnt1)/2*p;
89         }
90         long long tot = (b-a+1)*(d-c+1);
91         long long GCD = gcd(ans,tot);
92         ans /= GCD;
93         tot /= GCD;
94         printf("Case #%d: %I64d/%I64d\n",iCase,ans,tot);
95     }
96     return 0;
97 }

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

posted on 2013-11-18 11:27 kuangbin 阅读(...) 评论(...) 编辑 收藏

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