HDU 5475 An easy problem

An easy problem

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 594    Accepted Submission(s): 343

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

110 10000000001 22 11 21 102 32 41 61 71 122 7

 

Sample Output

Case #1:2122010164250484

 

Source

2015 ACM/ICPC Asia Regional Shanghai Online

 

裸的线段树。

单点更新。

#include<stdio.h>#include<iostream>#define lson l,m,rt<<1#define rson m+1,r,rt<<1 | 1using namespace std;const int maxn=100005;const int inf=1<<29;int n,m,t;long long sum[500000],M;void pushup(int rt){    sum[rt]=sum[rt<<1]*sum[rt<<1|1]%M;}void build(int l,int r,int rt){    if(l==r)    {        sum[rt]=1;        return ;    }    int m=(l+r)>>1;    build(lson);    build(rson);    pushup(rt);}void update(int i,int add,int l,int r,int rt){    if(l==r)    {        sum[rt]=add;        return ;    }    int m=(l+r)>>1;    if(i<=m)        update(i,add,lson);    else        update(i,add,rson);    pushup(rt);}long long query(int L,int R,int l,int r,int rt){    if(L<=l&&r<=R)    {        return sum[rt];    }    int m=(l+r)>>1;    long long res=1;    if(L<=m)        res=(res*query(L,R,lson))%M;    if(R>m)        res=(res*query(L,R,rson))%M;    return res;}int main(){    ios::sync_with_stdio(false);    cin>>t;    for(int p=1;p<=t;p++)    {        cout<<"Case #"<<p<<":"<<endl;        cin>>n>>M;        build(1,n,1);        for(int i=1;i<=n;i++)        {            int op,x;            cin>>op>>x;            if(op==1)            {                update(i,x,1,n,1);                cout<<query(1,i,1,n,1)<<endl;            }            else            {                update(x,1,1,n,1);                cout<<query(1,i,1,n,1)<<endl;            }        }    }    return 0;}