HDOJ 4790 Just Random

http://www.cnblogs.com/xin-hua/p/3553045.html

思路：对于a<=x<=b,c<=y<=d，满足条件的结果为ans=f(b,d)-f(b,c-1)-f(a-1,d)+f(a-1,c-1)。

而函数f(a,b)是计算0<=x<=a,0<=y<=b满足条件的结果。这样计算就很方便了。

例如：求f(16,7),p=6,m=2.

对于x有：0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4

对于y有：0 1 2 3 4 5 0 1

很容易知道对于xy中的（0 1 2 3 4 5）对满足条件的数目为p。

这样取A集合为（0 1 2 3 4 5 0 1 2 3 4 5），B集合为（0 1 2 3 4）。

C集合为（0 1 2 3 4 5），D集合为（0 1）。

这样就可以分成4部分来计算了。

Just Random

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1151    Accepted Submission(s): 315

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 <= 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
   
   
   
   

 

Source

2013 Asia Chengdu Regional Contest

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long int LL;

LL a,b,c,d,p,m;

LL getF(LL x,LL y)
{
    x++,y++;

    LL nn=x/p,mm=y/p;
    LL rn=x%p,rm=y%p;

    LL res1=nn*mm*p;
    LL res2=nn*rm;
    LL res3=mm*rn;
    LL res4=0;

    if(rn<=rm)
    {
        LL l1=rn;
        LL l2=rm;
        for(LL i=m+1;i<rn+rm;i+=p)
        {
            if(i<l1) res4+=i;
            else if(i>=l1&&i<=l2) res4+=rn;
            else res4+=rn+rm-i;
        }
    }
    else
    {
        LL l1=rm;
        LL l2=rn;
        for(LL i=m+1;i<rn+rm;i+=p)
        {
            if(i<l1) res4+=i;
            else if(i>=l1&&i<=l2) res4+=rm;
            else res4+=rn+rm-i;
        }
    }
    return res1+res2+res3+res4;
}

LL gcd(LL a,LL b)
{
    if(b==0) return a;
    return gcd(b,a%b);
}

int main()
{
	int T_T,cas=1;
	scanf("%d",&T_T);
	while(T_T--)
	{
		cin>>a>>b>>c>>d>>p>>m;
        LL A=(b-a+1)*(d-c+1);
        LL B=getF(b,d)-getF(a-1,d)-getF(b,c-1)+getF(a-1,c-1);
        LL G=gcd(A,B);
        cout<<"Case #"<<cas++<<": "<<B/G<<"/"<<A/G<<endl;
    }
    return 0;
}