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
题目主要是出现的除法，在模条件下是不能进行除法的，除非存在逆元可以实现除法，但是此处除数不一定与被除数互质。
但是如果过程中不模的话，就要使用大数，会T。
考虑到题目中提到了，除数不会出现相同的。
也就是如果乘了1,2,3,然后再除掉2的话，结果就是由1和3构成，这样就不用考虑每个数的情况了，此时的每个数就是一个整体，结果只和这个数有没有出现有关。
于是可以考虑用线段树来维护分段的积。当某一个数被除掉了，所有与这个数相关的区间都要重新计算，最多有log(q)个区间。
这样效率就是qlogq，是满足条件的。
代码：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <string>
#define LL long long
using namespace std;
const int maxN = 100005;
int q, m;
int op[maxN], top;
//线段树
struct node
{
int lt, rt;
LL val;
}tree[4*maxN];
//向上更新
void pushUp(int id)
{
tree[id].val = (tree[id<<1].val*tree[id<<1|1].val)%m;
}
//建立线段树
void build(int lt, int rt, int id)
{
tree[id].lt = lt;
tree[id].rt = rt;
tree[id].val = 1;//每段的初值，根据题目要求
if (lt == rt)
{
//tree[id].add = ??;
return;
}
int mid = (lt+rt)>>1;
build(lt, mid, id<<1);
build(mid+1, rt, id<<1|1);
pushUp(id);
}
void add(int lt, int rt, int id, int pls)
{
if (lt <= tree[id].lt && rt >= tree[id].rt)
{
if (pls)
{
tree[id].val *= pls;
tree[id].val %= m;
}
else
tree[id].val = 1;
return;
}
int mid = (tree[id].lt+tree[id].rt)>>1;
if (lt <= mid)
add(lt, rt, id<<1, pls);
if (rt > mid)
add(lt, rt, id<<1|1, pls);
pushUp(id);
}
void work()
{
build(1, q, 1);
top = 1;
int d, y;
for (int i = 0; i < q; ++i)
{
scanf("%d%d", &d, &y);
if (d == 1)
add(top, top, 1, y);
else
add(y, y, 1, 0);
op[top++] = y;
printf("%I64d\n", tree[1].val);
}
}
int main()
{
//freopen("test.in", "r", stdin);
int T;
scanf("%d", &T);
for (int times = 1; times <= T; ++times)
{
printf("Case #%d:\n", times);
scanf("%d%d", &q, &m);
work();
}
return 0;
}