HDU5475

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

 //2016.9.12
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #define N 100005

 using namespace std;

 int nu[N], book[N];

 int main()
 {
     long long ans;
     int T, kase = , q, mod, op;
     scanf("%d", &T);
     while(T--)
     {
         ans = ;
         memset(book, true, sizeof(book));
         printf("Case #%d:\n", ++kase);
         scanf("%d%d", &q, &mod);
         for(int i = ; i <= q; i++)
         {
             scanf("%d%d", &op, &nu[i]);
             if(op == )
             {
                 ans *= nu[i];
                 ans %= mod;
             }
             else
             {
                 book[nu[i]] = false;
                 book[i] = false;
                 ans = ;
                 for(int  j = ; j < i; j++)
                 {
                     if(book[j])ans = (ans*nu[j])%mod;
                 }
             }
             printf("%lld\n", ans);
         }
     }

     return ;
 }