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.
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).
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
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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
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;
}