An easy problem

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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

思路：线段树单点更新；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
const int N=1e5+10,M=4e6+10,inf=1e9+10;
ll sum[N<<2],mod;
void pushup(int pos)
{
    sum[pos]=(sum[pos<<1|1]*sum[pos<<1])%mod;
}
void buildtree(int l,int r,int pos)
{
    if(l==r)
    {
        sum[pos]=1;
        return;
    }
    int mid=(l+r)>>1;
    buildtree(l,mid,pos<<1);
    buildtree(mid+1,r,pos<<1|1);
    pushup(pos);
}
void update(int point,ll change,int l,int r,int pos)
{
    if(l==r&&l==point)
    {
        sum[pos]=change;
        return;
    }
    int mid=(l+r)>>1;
    if(point<=mid)
    update(point,change,l,mid,pos<<1);
    else
    update(point,change,mid+1,r,pos<<1|1);
    pushup(pos);
}
int main()
{
    int T,cas=1;
    scanf("%d",&T);
    while(T--)
    {
        int n,m;
        scanf("%d%lld",&n,&mod);
        buildtree(1,n,1);
        printf("Case #%d:\n",cas++);
        for(int i=1;i<=n;i++)
        {
            int flag;
            ll l;
            scanf("%d%lld",&flag,&l);
            if(flag==1)
            update(i,l,1,n,1);
            else
            update(l,1,1,n,1);
            printf("%lld\n",sum[1]);
        }
    }
    return 0;
}