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
题意：
初始X=1，有两种操作:
1.X乘上一个整数y。
2.X除以一个之前乘过的数。
要求每次操作之后，输出X%M的值
思路：
这道题咋一看好像是个简单的模拟题，但是仔细一想会发现除法直接除肯定不行（因为有取余操作啊）。那么最正确的做法就是将每次乘的数存起来，输出结果时直接把所有存的数乘起来就行了。而因为除的都是之前乘过的数，所以只需要把存的对应位置的乘数变成1就行了，这样就把除法转换成了乘法，进而排除了取余操作的影响。
当然这样做的时间复杂度是O（n^2）,但我们可以通过线段树维护每次操作的数，这样时间复杂度就降到了O（nlogn）。
另外这题数据可能比较水，O（n^2）其实直接就能过。
代码：
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN = 100005;
long long Tree[4*MAXN],MOD;
void Up(int temp){
Tree[temp] = (Tree[temp<<1]*Tree[temp<<1|1])%MOD;
}
void Build(int temp,int left,int right){
Tree[temp] = 1;
if(left == right)return ;
int m = left + (right-left)/2;
Build(temp<<1,left,m);
Build(temp<<1|1,m+1,right);
}
void Updata(int temp,int left,int right,int tempt,int value){
if(left == right){
Tree[temp] = value;
return ;
}
int m = left + (right-left)/2;
if(tempt<=m)Updata(temp<<1,left,m,tempt,value);
else Updata(temp<<1|1,m+1,right,tempt,value);
Up(temp);
}
int main(){
int T,N;
scanf("%d",&T);
for(int _=1 ; _<=T ; ++_){
printf("Case #%d:\n",_);
scanf("%d %d",&N,&MOD);
Build(1,1,N);
for(int i=1 ; i<=N ; ++i){
int a,b;
scanf("%d %d",&a,&b);
if(a == 1){
Updata(1,1,N,i,b);
printf("%lld\n",Tree[1]);
}
else {
Updata(1,1,N,b,1);
printf("%lld\n",Tree[1]);
}
}
}
return 0;
}