CF1842H.Tenzing and Random Real Numbers
普及/提高-
通过率:0%
AC君温馨提醒
该题目为【codeforces】题库的题目,您提交的代码将被提交至codeforces进行远程评测,并由ACGO抓取测评结果后进行展示。由于远程测评的测评机由其他平台提供,我们无法保证该服务的稳定性,若提交后无反应,请等待一段时间后再进行重试。
题目描述
There are n uniform random real variables between 0 and 1, inclusive, which are denoted as x1,x2,…,xn .
Tenzing has m conditions. Each condition has the form of xi+xj≤1 or xi+xj≥1 .
Tenzing wants to know the probability that all the conditions are satisfied, modulo 998 244 353 .
Formally, let M=998 244 353 . It can be shown that the answer can be expressed as an irreducible fraction qp , where p and q are integers and q≡0(modM) . Output the integer equal to p⋅q−1modM . In other words, output the integer x that 0≤x<M and x⋅q≡p(modM) .
输入格式
The first line contains two integers n and m ( 1≤n≤20 , 0≤m≤n2+n ).
Then following m lines of input contains three integers t , i and j ( t∈{0,1} , 1≤i≤j≤n ).
- If t=0 , the condition is xi+xj≤1 .
- If t=1 , the condition is xi+xj≥1 .
It is guaranteed that all conditions are pairwise distinct.
输出格式
Output the probability that all the conditions are satisfied, modulo M=998 244 353 .
输入输出样例
输入#1
3 2 0 1 2 1 3 3
输出#1
748683265
输入#2
3 3 0 1 2 0 1 3 0 2 3
输出#2
748683265
输入#3
3 4 0 1 2 0 1 3 1 2 3 1 2 2
输出#3
935854081
输入#4
4 4 0 1 2 0 3 4 1 1 3 1 2 4
输出#4
0
输入#5
8 12 0 1 2 0 2 3 1 3 4 0 1 4 0 5 6 0 6 7 1 7 8 0 5 8 1 3 7 1 3 8 1 4 7 1 4 8
输出#5
997687297
说明/提示
In the first test case, the conditions are x1+x2≤1 and x3+x3≥1 , and the probability that each condition is satisfied is 21 , so the probability that they are both satisfied is 21⋅21=41 , modulo 998 244 353 is equal to 748683265 .
In the second test case, the answer is 41 .
In the third test case, the answer is 161 .
In the fourth test case, the sum of all variables must equal 2 , so the probability is 0 .