题干：

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≤101≤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≤1091≤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≤1090<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

题目大意：

初始X=1；

给n，mod   ，表示n次操作

操作1格式 ： 1  b 表示用X乘上b

操作2格式： 2   n 表示 当前X除掉第n次 "1操作” 的数

每次操作都要输出一个答案，输出的答案是要对mod取模

解题报告：

由于有除法所以我们不能每一步取模,因为对于每一个数对mod并不是都存在逆元。

考虑线段树维护乘积，下标代表第几次操作，所以直接做就可以了。对于第i次操作，如果是1 b操作就是在下标为i的地方更新为b，如果是2 n 操作就是在下标是n的地方更新值为1，线段树维护区间积就可以了。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define F first
#define S second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 4e5 + 5;
ll mod;
struct TREE {
	int l,r;
	ll val;
} tr[MAX];
void pushup(int cur) {
	tr[cur].val = tr[cur*2].val * tr[cur*2+1].val % mod;
}
void build(int l,int r,int cur) {
	tr[cur].l = l;tr[cur].r = r;
	tr[cur].val = 1;
	if(l == r) {
		return;
	}
	int m = (l+r)>>1;
	build(l,m,cur*2);
	build(m+1,r,cur*2+1);
	pushup(cur); 
}
void update(int tar,int cur,ll val) {
	if(tr[cur].l == tr[cur].r) {
		tr[cur].val = val;
		return;
	}
	if(tar <= tr[cur*2].r) update(tar,cur*2,val);
	else update(tar,cur*2+1,val);
	pushup(cur);
}
int main()
{
	int t,iCase=0,op,m;
	ll x;
	cin>>t;
	while(t--) {
		scanf("%d%lld",&m,&mod);
		build(1,m,1);
		printf("Case #%d:\n",++iCase);
		for(int i = 1; i<=m; i++) {
			scanf("%d%lld",&op,&x);
			if(op == 1) {
				update(i,1,x);
				printf("%lld\n",tr[1].val%mod);
			}
			else {
				update(x,1,1);
				printf("%lld\n",tr[1].val%mod);
			}
		}
	}
	return 0 ;
}