An easy problem

An easy problem

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 216    Accepted Submission(s): 111

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
#include <iostream>
#include<stdio.h>
#include<malloc.h>
#include<string.h>
using namespace std;
long long y[111111];
long long x[111111];
long long mod;
bool vis[111111];
void cal(int st,int id)
{
  for(int i=st;i<=id;i++)    
  {
    x[i]=x[i-1];
    if(vis[i]) continue;
    x[i]=(x[i-1]*y[i])%mod;
  }
}
int main()
{
  int t,q,opr;
  scanf("%d",&t);
  for(int z=1;z<=t;z++)
  {
    memset(vis,0,sizeof(vis));
    scanf("%d %I64d",&q,&mod);
    x[0]=1;
    printf("Case #%d:\n",z);
    for(int i=1;i<=q;i++)
    {
      scanf("%d %I64d",&opr,&y[i]);
      if(opr==1)
      {
        x[i]=(x[i-1]*y[i])%mod;
        printf("%I64d\n",x[i]);
      }
      else
      {
        vis[y[i]]=1;
        vis[i]=1;
        cal(y[i],i);
        printf("%I64d\n",x[i]);
      }
    }
  }
  return 0;
}