Just Random
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2462 Accepted Submission(s): 717
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
寒假训练的一道题，在比赛中，对这种数学题格外情有独钟，两个小时一直在想这个题，当有思路时已经没时间了，比赛结束又是各种WA，主要原因在于 边界没有处理好！
题目大意：
给你两个闭区间：[a,b],[c,d]，让你在这两个区间内各选一个数，x，y，问取到的数满足(x+y)mod p = m的概率！
分析：
这个题大体思路很好想：那就是找到符合条件的，除以总情况， 总情况很好找：就是(b-a+1) * (d-c+1)；
关键是符合条件的：
可以给(x+y)mod p = m变一下型，变成 n * p +m = ( x + y );这里n是整数。这里n相当于整体解！就是余数相同时，有几个结果！
在变换一下： y = -x + (n*p + m) 很明显 这是一次函数，只有p是变量，因此他是一个上下平移的直线！
很明显 总情况就是 y∈[ c , d ] ，x ∈ [ a , b ]的由点构成的矩形，就是求直线与矩形的交点的个数。这就是所有符合条件的解！
一开始我的做法是直接循环扫描 就是一条直线一条直线的扫，结果超时！因此，这个题目必须把边界球出来，直接用数学算出来 不能循环！
边界分析：
很明显 四条直线把整个矩形分成三个区域，而求交点，显然A1是递减的等差数列，A2是常数列（全都是b-a+1个交点），A3是递增的等差数列！
只需要求出n1 n2 n3 n4，边界在求数列和即可！
需要注意的是这个图是适合c-d > b-a的情况， 对于b-a > d-c的情况可以把bd ca交换，相当于y ∈[a,b]不管怎样变换 只是形式的变换 结果不可能变得 直接swap即可！
这里我错了很多次，是因为边界没处理好，就是有重复计算！
可以把没一个区域具体化！就是A1有 n11 和 n12 构成 A2 由 n21 n22 构成 A3 由 n31 n32 构成 ！
不管怎样总有一个适合自己的方法！
最后还要除以最大公约数写个gcd（）即可！
代码如下：
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b){
if (!b)return a;
return gcd(b,a%b);
}
ll a,b,c,d,p,m;
ll f(ll x,ll n){
return -x+(n*p+m);
}
int main()
{
ll n;
scanf("%lld",&n);
for (ll cont = 1;cont <= n; ++cont){
scanf("%lld%lld%lld%lld%lld%lld",&a,&b,&c,&d,&p,&m);
ll n11,n12,n21,n22,n31,n32;
if (b-a > d-c){
swap(b,d);
swap(c,a);
}
n11 = ceil((double)(a+c-m)/p);
n12 = (b+c-m )% p == 0 ? (b+c-m) / p - 1 : floor((double)(b+c-m)/p);// == n1;
n21 = (b+c-m )% p == 0 ? (b+c-m) / p : ceil((double)(b+c-m)/p);
n22 = (a+d-m)%p == 0 ? (a+d-m) / p : floor((double)(a+d-m)/p);
n31 = (a+d-m)%p == 0 ? (a+d-m) / p + 1: ceil((double)(a+d-m)/p);
n32 = (b + d - m )% p == 0 ? (b+d-m)/p : floor((double)(b+d-m)/p);
ll sum=0;
sum += (n22 - n21 + 1) * (b - a + 1);
ll b2 = n11 * p + m - c;
ll st1 = b2 - a + 1;
ll num1 = n12 - n11 + 1;
sum += num1*st1 + (num1-1)*num1*p / 2;
ll a2 = n32 * p + m - d;
ll st2 = b - a2 + 1;
ll num2 = n32-n31+ 1;
sum+=num2*st2+num2*(num2-1)*p/2;
ll len=(d-c+1)*(b-a+1);
ll gd=gcd(len,sum);
printf("Case #%lld: %lld/%lld\n",cont,sum/gd,len/gd);
}
return 0;
}
