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 == 0)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 = 0;
    scanf("%d",&T);
    while(T--)
    {
        iCase++;
        scanf("%I64d%I64d%I64d%I64d%I64d%I64d",&a,&b,&c,&d,&p,&m);
        long long ans = 0;
        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-1)%p;
            long long sub = (t2 - m + p)%p;
            long long cnt2 = (b+c-1-sub-m)/p;
            //cout<<t2<<" "<<sub<<endl;
            ans += (cnt2 - cnt1 + 1)*(1+add) + (cnt2 - cnt1 + 1)*(cnt2 - cnt1)/2 * 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 + 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;
            ans += (cnt2 - cnt1 + 1)*(1+sub) + (cnt2 - cnt1 + 1)*(cnt2 - cnt1)/2*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-1)%p;
            long long sub = (t2 - m + p)%p;
            long long cnt2 = (a+d-1-sub-m)/p;
            ans += (cnt2 - cnt1 + 1)*(1+add) + (cnt2 - cnt1 + 1)*(cnt2 - cnt1)/2 * 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 + 1)*(d-c+1);
            t1 = (b+c+1)%p;
            add = (m - t1 + p)%p;
            cnt1 = (b+c+1+add-m)/p;
            t2 = (b+d)%p;
            sub = (t2 - m + p)%p;
            cnt2 = (b+d - sub-m)/p;
            ans += (cnt2 - cnt1 + 1)*(1+sub) + (cnt2 - cnt1 + 1)*(cnt2 - cnt1)/2*p;
        }
        long long tot = (b-a+1)*(d-c+1);
        long long GCD = gcd(ans,tot);
        ans /= GCD;
        tot /= GCD;
        printf("Case #%d: %I64d/%I64d\n",iCase,ans,tot);
    }
    return 0;
}