hdu5475 An easy problem(好题)

An easy problem

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 113    Accepted Submission(s): 55

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


比赛的时候一直以为是求逆元,结果卡了几个小时,
题目就是建立一个简单的线段树,然后单点修改求区间

#include <cstdio>
#include <cstring>
#include <algorithm>
#include  <iostream>
using namespace std;
typedef long long ll;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int maxn=155555;
ll sum[maxn*4+4];
int add;
int a[maxn];
int L,R;
int MOD;
int p;

void pushup(int rt){
    sum[rt]=sum[rt<<1]*sum[rt<<1|1]%MOD;
}

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 l,int r,int rt){
    if(l==r){
        sum[rt]=add;
        return ;
    }
    int m=(l+r)>>1;
    if(p<=m)
        update(lson);
    else
        update(rson);
    pushup(rt);
}

ll query(int l,int r,int rt){
    ll ret=1;
    if(L<=l&&R>=r)
        return sum[rt];
    int m=(l+r)>>1;
    if(L<=m)
        ret*=query(lson)%MOD;
    if(R>m)
        ret*=query(rson)%MOD;
    return ret%MOD;
}

int main()
{
    int n,m;
    int op,u,q;
    int t;
    scanf("%d",&t);
    for(int case1=1;case1<=t;case1++){
        printf("Case #%d:\n",case1);
        scanf("%d%d",&q,&MOD);
        build(1,q,1);
        for(int i=1;i<=q;i++){
            scanf("%d%d",&op,&u);
            if(op==1){
                p=i;
                add=u;
                update(1,q,1);
                L=1;
                R=i;
                printf("%I64d\n",query(1,q,1)%MOD);
            }
            else{
                p=u;
                add=1;
                update(1,q,1);
                L=1;
                R=i;
                printf("%I64d\n",query(1,q,1)%MOD);
            }
        }
    }
    return 0;
}