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 = 0, q, mod, op;
    scanf("%d", &T);
    while(T--)
    {
        ans = 1;
        memset(book, true, sizeof(book));
        printf("Case #%d:
", ++kase);
        scanf("%d%d", &q, &mod);
        for(int i = 1; i <= q; i++)
        {
            scanf("%d%d", &op, &nu[i]);
            if(op == 1)
            {
                ans *= nu[i];
                ans %= mod;
            }
            else 
            {
                book[nu[i]] = false;
                book[i] = false;
                ans = 1;
                for(int  j = 1; j < i; j++)
                {
                    if(book[j])ans = (ans*nu[j])%mod;
                }
            }
            printf("%lld
", ans);
        }
    }

    return 0;
}