hdu5475 An easy problem

An easy problem

Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 873 Accepted Submission(s): 472

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 

题意：两种操作，当x=1时，乘y，当x=2时，除以第y个操作乘的数，输出结果，对M取模。（注意只有输出的时候才取模）

分析：网赛的时候没写出来，老是想着怎么用逆元处理除法取余，但就是搞不定，之后看到一篇博客，想法很不错，虽然时间久点，但题目给了5S，也还是能过的；具体就是把除法也看做乘法，也就是当做至始至终没乘过要除的那个数。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a) memset(a,0,sizeof(a))

ll a[100010],vis[100010];//a[]存所有的乘数，vis[]标记这个数之后是否要被除
ll Q,M;
ll x,y,ans;

int main ()
{
    int T;
    scanf ("%d",&T);
    for (int cas=1; cas<=T; cas++)
    {
        CL(vis);
        scanf ("%lld%lld",&Q,&M);
        printf ("Case #%d:\n",cas);
        ans = 1;
        for (int i=1; i<=Q; i++)
        {
            scanf ("%lld%lld",&x,&y);
            if (x==1)//乘法直接乘
            {
                vis[i] = 1;
                ans = (ans*y)%M;
                a[i] = y;
            }
            if (x==2)//除法就不乘要除的数
            {
                ans = 1;
                for (int j=1; j<i; j++)
                {
                    if (j == y) vis[j] = 0;//标记这个数需要除了，之后就不用再乘了
                    else if (vis[j]) ans = (ans*a[j])%M;
                }
                a[i] = 1;//这部好像可以不要，但是也没影响
            }
            printf ("%lld\n",ans);
        }
    }
    return 0;
}