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 给出 a<=x<=b c<=y<=d 求满足 （x+y） % p ==m 的概率这题需要找规律 然后根据规律求解 每一个数出现的次数为 上升的等差数列 相同的值 下降的等差数列 以下看规律
1 0 5 0 5 3 0 2 0 1 1 2 2 2 3 3 3 3 4 4 4 4 4 5 5 5 5 5 5 6 6 6 6 6 7 7 7 7 8 8 8 9 9 10 3 3 7 2 5 4 6 4 5 6 6 7 7 7 8 8 8 8 9 9 9 9 10 10 10 11 11 12 5 1 3 2 6 2 5 6 3 4 4 5 5 5 6 6 6 7 7 7 8 8 9
然后根据等差数列求和 求解
1 #include <cstdio>
2 #include <cstring>
3 #include <queue>
4 #include <cmath>
5 #include <algorithm>
6 #include <set>
7 #include <iostream>
8 #include <map>
9 #include <stack>
10 #include <string>
11 #include <vector>
12 #define pi acos(-1.0)
13 #define eps 1e-6
14 #define fi first
15 #define se second
16 #define lson l,m,rt<<1
17 #define rson m+1,r,rt<<1|1
18 #define rtl rt<<1
19 #define rtr rt<<1|1
20 #define bug printf("******\n")
21 #define mem(a,b) memset(a,b,sizeof(a))
22 #define name2str(x) #x
23 #define fuck(x) cout<<#x" = "<<x<<endl
24 #define f(a) a*a
25 #define sf(n) scanf("%d", &n)
26 #define sff(a,b) scanf("%d %d", &a, &b)
27 #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
28 #define sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
29 #define pf printf
30 #define FRE(i,a,b) for(i = a; i <= b; i++)
31 #define FREE(i,a,b) for(i = a; i >= b; i--)
32 #define FRL(i,a,b) for(i = a; i < b; i++)
33 #define FRLL(i,a,b) for(i = a; i > b; i--)
34 #define FIN freopen("in.txt","r",stdin)
35 #define gcd(a,b) __gcd(a,b)
36 #define lowbit(x) x&-x
37 using namespace std;
38 typedef long long LL;
39 typedef unsigned long long ULL;
40 const int mod = 1e9 + 7;
41 const int maxn = 1e5 + 10;
42 const int INF = 0x3f3f3f3f;
43 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
44 int _;
45 LL a, b, c, d, m, p;
46 LL solveL ( LL L, LL R ) {
47 LL cnt = L % p;
48 if ( cnt > m ) cnt = L + p - cnt + m;
49 else cnt = L + m - cnt ;
50 if ( cnt > R ) return 0;
51 else {
52 int temp = cnt - ( a + c ) + 1;
53 int num = ( R - cnt ) / p + 1;
54 LL sum = 1LL * temp * num + 1LL * num * ( num - 1 ) * p / 2;
55 return sum;
56 }
57 }
58 LL solveM ( LL L, LL R ) {
59 LL cnt = L % p;
60 if ( cnt > m ) cnt = L + p - cnt + m;
61 else cnt = L + m - cnt ;
62 if ( cnt > R ) return 0 ;
63 else {
64 LL temp = L - ( a + c ) + 1;
65 LL num = ( R - cnt ) / p + 1;
66 LL sum = 1LL * temp * num;
67 return sum;
68 }
69 }
70 LL solveR ( LL L, LL R ) {
71 LL cnt = L % p;
72 if ( cnt > m ) cnt = L + p - cnt + m;
73 else cnt = L + m - cnt ;
74 if ( cnt > R ) return 0 ;
75 else {
76 LL temp = ( b + d ) - cnt + 1;
77 LL num = ( R - cnt ) / p + 1;
78 LL sum = 1LL * temp * num - 1LL * num * ( num - 1 ) * p / 2;
79 return sum;
80 }
81 }
82 int main() {
83 sf ( _ );
84 int cas = 1;
85 while ( _-- ) {
86 scanf ( "%lld%lld%lld%lld%lld%lld", &a, &b, &c, &d, &p, &m );
87 LL L = a + c, R = b + d, l = min ( b + c, a + d ), r = max ( b + c, a + d ), ans = 0;
88 ans += solveL ( L, l - 1 );
89 ans += solveM ( l, r );
90 ans += solveR ( r + 1, R );
91 LL sum = ( b - a + 1 ) * ( d - c + 1 );
92 LL g = gcd ( ans, sum );
93 printf ( "Case #%d: %lld/%lld\n", cas++, ans / g, sum / g );
94 }
95 return 0;
96 }
原文地址：https://www.cnblogs.com/qldabiaoge/p/9703858.html