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

巴特西

HDU 5475An easy problem 离线set/线段树

An easy problem

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