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.
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).
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).
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
#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;
}