An easy problem
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
题意：T组案例，输入n，MOD，以下n组操作，op，x，当op = 1时，M * = x（M初始值为1），当op = 2时，M除以第x次操作的结果。过程中每次变化都mod。输出每次操作后的结果。
思路：建立线段树，乘法操作的时候就乘，作除法操作时，只需要不乘第i次的数，每次都计算tree[1].sum即可。可以从线段树上设立标记，当进行除第i个数的操作时，就不乘这个位置。（线段树是真的牛！）
AC代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<utility>
#include<set>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#define maxn 100005
#define INF 0x3f3f3f3f
#define LL long long
#define ULL unsigned long long
#define E 1e-8
#define mod 1000000007
#define P pair<int,int>
#define MID(l,r) (l+(r-l)/2)
#define lson(o) (o<<1) //o*2
#define rson(o) (o<<1|1) //o*2+1
using namespace std;
int M;
int p,v;
struct node
{
int l,r;
LL sum;
bool flag;
}tree[maxn<<2];
void build(int o,int l,int r)
{
tree[o].l = l;
tree[o].r = r;
tree[o].flag = 1;
if(l == r){
tree[o].sum = 1;
return ;
}
int m = MID(l,r);
int lc = lson(o),rc = rson(o);
build(lc,l,m);
build(rc,m+1,r);
tree[o].sum = 1; //不能忽略
if(tree[lc].flag==1) tree[o].sum = tree[o].sum*tree[lc].sum % M;
if(tree[rc].flag==1) tree[o].sum = tree[o].sum*tree[rc].sum % M;
}
void update(int t,int o)
{
if(tree[o].l == tree[o].r){
if(t == 1){
tree[o].flag = 1;
tree[o].sum = v;
}
else{
tree[o].flag = 0;
}
return ;
}
int m = MID(tree[o].l,tree[o].r);
int lc = lson(o),rc = rson(o);
if(p <= m) update(t,lc);
else update(t,rc);
tree[o].sum = 1; //不能忽略
if(tree[lc].flag==1) tree[o].sum = tree[o].sum*tree[lc].sum % M;
if(tree[rc].flag==1) tree[o].sum = tree[o].sum*tree[rc].sum % M;
}
int main()
{
int T,cases = 0;
int n,temp,op,x;
scanf("%d",&T);
while(T--){
printf("Case #%d:\n",++cases);
scanf("%d %d",&n,&M);
build(1,1,n);
for(int i=1;i<=n;++i){
scanf("%d %d",&op,&x);
if(op == 1){
temp = 1;
p = i;
v = x;
}
else{
temp = 0;
p = x;
}
update(temp,1);
printf("%lld\n",tree[1].sum);
}
}
}