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HDU5475

2024-08-13 05:32:23   219人阅读

An easy problem

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1697    Accepted Submission(s): 760


Problem Description

One day, a useless calculator was being built by Kuros. Let‘s assume that number X is showed on the screen of calculator. At first, X = 1. This calculator only supports two types of operation.
1. multiply X with a number.
2. divide X with a number which was multiplied before.
After each operation, please output the number X modulo M.
 

 

Input

The first line is an integer T(1≤T≤10), indicating the number of test cases.
For each test case, the first line are two integers Q and M. Q is the number of operations and M is described above. (1≤Q≤105,1≤M≤109)
The next Q lines, each line starts with an integer x indicating the type of operation.
if x is 1, an integer y is given, indicating the number to multiply. (0<y≤109)
if x is 2, an integer n is given. The calculator will divide the number which is multiplied in the nth operation. (the nth operation must be a type 1 operation.)

It‘s guaranteed that in type 2 operation, there won‘t be two same n.
 

 

Output

For each test case, the first line, please output "Case #x:" and x is the id of the test cases starting from 1.
Then Q lines follow, each line please output an answer showed by the calculator.
 

 

Sample Input

1
10 1000000000
1 2
2 1
1 2
1 10
2 3
2 4
1 6
1 7
1 12
2 7
 

 

Sample Output

Case #1:
2
1
2
20
10
1
6
42
504
84
 

 

Source

2015 ACM/ICPC Asia Regional Shanghai Online
 
既然说是简单题，那就不用想的太复杂，暴力的做法也能过
 1 //2016.9.12 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #define N 100005 6  7 using namespace std; 8  9 int nu[N], book[N];10 11 int main()12 {13     long long ans;14     int T, kase = 0, q, mod, op;15     scanf("%d", &T);16     while(T--)17     {18         ans = 1;19         memset(book, true, sizeof(book));20         printf("Case #%d:\n", ++kase);21         scanf("%d%d", &q, &mod);22         for(int i = 1; i <= q; i++)23         {24             scanf("%d%d", &op, &nu[i]);25             if(op == 1)26             {27                 ans *= nu[i];28                 ans %= mod;29             }30             else 31             {32                 book[nu[i]] = false;33                 book[i] = false;34                 ans = 1;35                 for(int  j = 1; j < i; j++)36                 {37                     if(book[j])ans = (ans*nu[j])%mod;38                 }39             }40             printf("%lld\n", ans);41         }42     }43 44     return 0;45 }

 

 

HDU5475


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