HDOJ Just Random 4790【2013成都区域赛J题+等差数列优化】

Just Random

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2102    Accepted Submission(s): 592

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 <= 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).

 

Sample Input

40 5 0 5 3 00 999999 0 999999 1000000 00 3 0 3 8 73 3 4 4 7 0

 

Sample Output

Case #1: 1/3Case #2: 1/1000000Case #3: 0/1Case #4: 1/1

 

Source

2013 Asia Chengdu Regional Contest

 

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终于看懂这题的代码了。。。

斌神代码：点击打开链接

利用等差数列优化。。。等差数列公式：

等差数列前n项和公式

• 

计算第一个符合条件的a1  然后在题中d=p。

注意计算的位置  计算末项时记得算前面的。

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;typedef long long LL;LL gcd(LL a,LL b){if(b==0)return a;return gcd(b,a%b);}int main(){int T;scanf("%d",&T);int xp=1;while(T--){LL a,b,c,d,p,m;scanf("%lld%lld%lld%lld%lld%lld",&a,&b,&c,&d,&p,&m);if((a+d)<(b+c)){swap(a,c);swap(b,d);}LL ans=0;LL t1=(a+c)%p; LL add=(m-t1+p)%p;LL cnt1=(a+c+add-m)/p;//计算a+c后面的第一个符合要求的数LL t2=(b+c-1)%p;LL s=(t2-m+p)%p;//因为要计算前一个数所以要减去s，为了符合等式减去的是p-(m-t2). LL cnt2=(b+c-1-s-m)/p;//计算b+c前面的第一个符合要求的数ans+=(cnt2-cnt1+1)*(1+add)+(cnt2-cnt1+1)*(cnt2-cnt1)/2*p;t1=(b+c)%p;add=(m-t1+p)%p;cnt1=(b+c+add-m)/p;t2=(a+d)%p;s=(t2-m+p)%p;cnt2=(a+d-s-m)/p;ans+=(cnt2-cnt1+1)*(b-a+1);t1=(a+d+1)%p;add=(m-t1+p)%p;cnt1=(a+d+1+add-m)/p;t2=(b+d)%p;s=(t2-m+p)%p;cnt2=(b+d-s-m)/p;ans+=(cnt2-cnt1+1)*(1+s)+(cnt2-cnt1+1)*(cnt2-cnt1)/2*p;long long num=(b-a+1)*(d-c+1);LL GCD=gcd(ans,num);ans/=GCD;num/=GCD;printf("Case #%d: %lld/%lld\n",xp++,ans,num);}return 0;}