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:
. Coach Pang randomly choose a integer x in [a, b] with equal probability.
. Uncle Yang randomly choose a integer y in [c, d] with equal probability.
. If (x + y) mod p = m, they will go out and have a nice day together.
. 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( <= a <= b <= , <=c <= d <= , <= m < p <= ).
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
Sample Output
Case #: /
Case #: /
Case #: /
Case #: /
Source
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。
f(16,7)=A和C满足条件的数+A和D满足条件的数+B和C满足条件的数+B和D满足条件的数。
其中前3个很好求的，关键是B和D满足条件的怎么求！
这个要根据m来分情况。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define ll long long
ll a,b,c,d,p,m;
ll gcd(ll x,ll y)
{
return y==?x:gcd(y,x%y);
}
ll f(ll x,ll y)
{
if(x< || y<)
return ;
ll mx=x%p;
ll my=y%p;
ll ans=;
ans=ans+(x/p)*(y/p)*p;//
ans=ans+(x/p)*(my+); //
ans=ans+(y/p)*(mx+);// if(mx>m)//
{
ans=ans+min(my,m)+;
ll t=(p+m-mx);
if(t<=my)
ans=ans+my-t+;
}
else//
{
ll t=(p+m-mx)%p;
if(t<=my) ans=ans+min(my-t+,m-t+);
}
return ans;
}
int main()
{
int t;
int ac=;
scanf("%d",&t);
while(t--)
{
printf("Case #%d: ",++ac);
scanf("%I64d%I64d%I64d%I64d%I64d%I64d",&a,&b,&c,&d,&p,&m);
ll ans=f(b,d)-f(b,c-)-f(a-,d)+f(a-,c-);
ll tot=(b-a+)*(d-c+);
ll r=gcd(ans,tot);
printf("%I64d/%I64d\n",ans/r,tot/r);
}
return ;
}
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