HDU 5475 An easy problem

An easy problem

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

Total Submission(s): 594    Accepted Submission(s): 343
[/b]

[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
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.

 

[align=left]Sample Input[/align]

1
10 1000000000
1 2
2 1
1 2
1 10
2 3
2 4
1 6
1 7
1 12
2 7

 

[align=left]Sample Output[/align]

Case #1:
2
1
2
20
10
1
6
42
504
84

 

[align=left]Source[/align]
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 | 1
using 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;
}