HDU 5475An easy problem 离线set/线段树

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

 

 

题解：考虑到取摸，离线做

或者线段树上搞

///
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
#include<bitset>
#include<set>
#include<vector>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a));
#define TS printf("111111\n");
#define FOR(i,a,b) for( int i=a;i<=b;i++)
#define FORJ(i,a,b) for(int i=a;i>=b;i--)
#define READ(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define mod 1000000007
#define inf 100000
inline ll read()
{
    ll x=,f=;
    char ch=getchar();
    while(ch<''||ch>'')
    {
        if(ch=='-')f=-;
        ch=getchar();
    }
    while(ch>=''&&ch<='')
    {
        x=x*+ch-'';
        ch=getchar();
    }
    return x*f;
}
//****************************************
struct ss
{
    int id,x;
    bool operator < (const ss &A)const
    {
        return id < A.id;
    }
};
#define maxn 100000+5
set<ss >s;
set<ss >::iterator it;
int main()
{
    int oo=;
    ll n,q,m,x[maxn],op[maxn],vis[maxn],ans[maxn],A[maxn];
    int T=read();
    while(T--)
    {
        scanf("%I64d%I64d",&n,&m);
        FOR(i,,n)
        {
            scanf("%I64d%I64d",&op[i],&x[i]);
        }
        mem(vis);
        FOR(i,,n)
        {
            if(op[i]==)vis[x[i]]=;
        }
        mem(ans);
        ans[]=;
        FOR(i,,n)
        {
            ans[i]=ans[i-];
            if(op[i]==&&!vis[i])
            {
                ans[i]=(ans[i]*x[i])%m;
            }
        }
        //cout<<ans[10]<<endl;
        s.clear();
        for(int i=n; i>=; i--)
        {
            ll tmp=;
            for(it=s.begin(); it!=s.end(); it++)
            {
                if((*it).id>i)break;
                tmp*=(*it).x;
                tmp%=m;
            }
            A[i]=(ans[i]*tmp)%m;
            if(op[i]==)
            {
                ss g;
                g.id=x[i];
                g.x=x[x[i]];
                s.insert(g);
            }
        }
        printf("Case #%d:\n",oo++);
        for(int i=; i<=n; i++)
            cout<<A[i]<<endl;
    }
    return ;
}

代码