A+B Problem

Posted on 2026-02-06 18:40  贾普鲁登  阅读(2)  评论(0)    收藏  举报 
A-A+B Problem_2026牛客寒假算法基础集训营1 

1 #include <bits/stdc++.h>
 2 using namespace std;
 3 long long p[10],q[10],num[10];
 4 const long long m = 998244353;
 5 long long ksm(int a,int b)
 6 {
 7     long long t = a,ans = 1;
 8     while(b)
 9     {
10         if(b % 2 == 1) ans = ans * t % m;
11         t = t * t % m;
12         b /= 2;
13     }
14     return ans;
15 }
16 long long cal(long long n)
17 {
18     long long ans = 1;
19     if(n == 0) ans *= num[0]*num[0]%m*num[0]%m*num[0]%m;
20     else if(n > 0 && n <= 9) ans *= num[0]*num[0]%m*num[0]%m;
21     else if(n >= 10 && n <= 99) ans *= num[0]*num[0]%m;
22     else if(n >= 100 && n <= 999) ans *= num[0];
23     while(n)
24     {
25         ans *= num[n % 10];
26         n /= 10;
27         ans %= m;
28     }
29     return ans;
30 }
31 //a/b mod m -> a * pow(b,m-2) 逆元
32 //why (pi/100)^56 cannot work but pi^56/100^56 does work?
33 long long inv100 = ksm(ksm(100,56),m-2);
34 int main()
35 {
36     int T,number;
37     long long ans = 0;
38     cin >> T;
39     while(T--)
40     {
41         ans = 0;
42         cin >> number;
43         for(int i = 1;i <= 7;i++) 
44         {
45             cin >> p[i];
46             p[i] = p[i];
47             q[i] = 100 - p[i];
48         }
49         // cautious every bulb will be tried to fire
50         num[0] = p[1]*p[2]%m*p[3]%m*q[4]%m*p[5]%m*p[6]%m*p[7]%m;
51         num[1] = q[1]*q[2]%m*p[3]%m*q[4]%m*q[5]%m*p[6]%m*q[7]%m;
52         num[2] = p[1]*q[2]%m*p[3]%m*p[4]%m*p[5]%m*q[6]%m*p[7]%m;
53         num[3] = p[1]*q[2]%m*p[3]%m*p[4]%m*q[5]%m*p[6]%m*p[7]%m;
54         num[4] = q[1]*p[2]%m*p[3]%m*p[4]%m*q[5]%m*p[6]%m*q[7]%m;
55         num[5] = p[1]*p[2]%m*q[3]%m*p[4]%m*q[5]%m*p[6]%m*p[7]%m;
56         num[6] = p[1]*p[2]%m*q[3]%m*p[4]%m*p[5]%m*p[6]%m*p[7]%m;
57         num[7] = p[1]*q[2]%m*p[3]%m*q[4]%m*q[5]%m*p[6]%m*q[7]%m;
58         num[8] = p[1]*p[2]%m*p[3]%m*p[4]%m*p[5]%m*p[6]%m*p[7]%m;
59         num[9] = p[1]*p[2]%m*p[3]%m*p[4]%m*q[5]%m*p[6]%m*p[7]%m;
60         for(int i = 0;i <= number;i++)
61         {
62             //inv100 is (1/100)^56 mod m -> pow(pow(100,56),m-2)
63             ans = (ans + cal(i) * cal(number-i) % m * inv100 ) % m;
64         }
65         cout << ans % m << endl;
66     }
67 }