hdu 4790 Just Random题解（数学）

http://acm.hdu.edu.cn/showproblem.php?pid=4790

Just Random

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1034    Accepted Submission(s): 284

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  

题意：a <= x <= b,c <= y<= d,求满足 (x+y)%p==m 的（x，y）的个数。

题解：（x+y）%p==m, (x+y)==p*K+m。假设x 固定，那么对于区间【c,d】,满足条件的最大的Kmax为，Kmax=floor( (x + d-m) / p)。 floor为向下取整。

满足条件的最小的Kmin为：Kmin=ceil( (x+c-m)/p )。ceil为向上取整。所以当x固定是，满足条件的个数为： Kmax-Kmin+1。（(x+d-m)<0或(x+c-m)<0时同样满足该式，因为

m<p)。 因为 a<=x<=b ，所以容易求出 sum(Kmax)和 sum（Kmin）。那么答案就sum（Kmax）-sum（Kmin）+（b-a+1）。（具体见代码）

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<string>
#include<math.h>
#define nn 510
#define inff 0x3fffffff
using namespace std;
typedef __int64 LL;
LL a,b,c,d,p,m;
LL jie(LL x) //当x==d-m时求 Kmax 的和，当x==c-m+p-1 时求 Kmin的和 （因为向上取整可以转化为向下取整，所以只写了一个求和的函数）
{
    LL z=floor(1.0*(b+x)/p);
    LL zz=floor(1.0*(a+x)/p);
    if(zz==z)
        return z*(b-a+1);
    LL re=0 ;
    LL ix=(z-1)*p-x;
    ix+=p;
    re+=z*(b-ix+1);
    ix=(zz+1)*p-x;
    ix--;
    re+=zz*(ix-a+1);
    LL f1=zz+1,f2=z  -1;
    if(f1<=f2)
        re+=(f1+f2)*(f2-f1+1)/2*p;
    return re;
}
LL solve()
{
    return jie(d-m)-jie(c-m+p-1)+b-a+1;
}
LL gcd(LL x,LL y)
{
    if(y==0)
        return x;
    return gcd(y,x%y);
}
int main()
{
    int t,cas=1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d%I64d%I64d%I64d%I64d%I64d",&a,&b,&c,&d,&p,&m);
        LL fz=solve();
        LL fm=(b-a+1)*(d-c+1);
        LL ix=gcd(fz,fm);
        printf("Case #%d: ",cas++);
        printf("%I64d/%I64d\n",fz/ix,fm/ix);
    }
    return 0;
}