hud 4790

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Just Random

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1123    Accepted Submission(s): 308

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

 

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>#define ll long longusing namespace std;ll maxn, minn;ll tip;ll help(ll a){    if ((a<minn)||(a>maxn))        return 0;    else        return min(min(a-minn, maxn-a)+1, tip);}ll gcd(ll a, ll b){    if (b==0)        return a;    else return gcd(b, a%b);}int main(){    //freopen("in.txt", "r", stdin);    int t;    scanf("%d", &t);    int cc = 0;    while (t--)    {        cc++;        ll a, b, c, d, p, m;        scanf("%I64d %I64d %I64d %I64d %I64d %I64d", &a, &b, &c, &d, &p, &m);         minn = a+c;        maxn = b+d;        tip = min(b-a, d-c)+1;        ll tip1 = minn+tip-1;        ll tip2 = maxn-tip+1;        ll mp1 = max((ll)0, (tip1-m)/p);        ll mp2 = max((ll)0, (tip2-m)/p);                ll fp = max((ll)0, (minn-m-1)/p);        if ((m+fp*p)<minn)            fp++;        ll lp = max((ll)0, (maxn-m)/p);        ll ans1=0;        ll tmp1 = help(m+fp*p);        ll tmp2 = help(m+lp*p);                if (mp1<fp)            mp1 = fp-1;        else            ans1 += (2*tmp1+(mp1-fp)*p)*(mp1-fp+1)/2;                if (lp<=mp2)            mp2 = lp;        else            ans1 += (2*tmp2+(lp-mp2-1)*p)*(lp-mp2)/2;               ans1 += (mp2-mp1)*tip;        ll ans2 = (b-a+1)*(d-c+1);        ll gd  = gcd(max(ans1, ans2), min(ans1, ans2));        printf("Case #%d: ", cc);        printf("%I64d/%I64d\n", ans1/gd, ans2/gd);    }    return 0;}