HDU 5475（2015 ICPC上海站网络赛）--- An easy problem（线段树点修改）

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=5475

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

 

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题意：输入Q和M，然后输入Q次操作，每次操作为一行数据op和s，定义一个数x=1。如果op=1表示x=x*s%mod 输出x，如果op=2 表示x除以第s次操作的s（第s次操作的op一定为1），输出x;

 

思路：线段树点修改，每次操作时，修改那个点，复杂度为log（n）,如果op=1，那么修改为s，如果op=2，那么第s次操作对应的点修改为1；

 

代码如下：

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <cstring>
#include <cmath>
#include <map>
#include <bitset>
using namespace std;
typedef long long LL;
LL mod;
LL tr[4*100005];
int p;
LL add;
void build(int l,int r,int i)
{
    if(l==r) {
        tr[i]=1;
        return ;
    }
    int mid=(l+r)>>1;
    build(l,mid,i<<1);
    build(mid+1,r,i<<1|1);
    tr[i]=1;
}

void update(int l,int r,int i)
{
    if(l==r) { tr[i]=add; return ; }
    int mid=(l+r)>>1;
    if(p<=mid) update(l,mid,i<<1);
    else update(mid+1,r,i<<1|1);
    tr[i]=(tr[i<<1]*tr[i<<1|1])%mod;
}

int main()
{
    int T,Q,Case=1;
    cin>>T;
    while(T--)
    {
        scanf("%d%lld",&Q,&mod);
        printf("Case #%d:
",Case++);
        build(1,Q,1);
        for(int i=1;i<=Q;i++)
        {
            int x;
            LL y;
            scanf("%d%lld",&x,&y);
            if(x==1) { p=i; add=y; }
            else { p=y; add=1; }
            update(1,Q,1);
            printf("%lld
",tr[1]);
        }
    }
    return 0;
}