【线段树】单点更新 hdu 5475 An easy problem

An easy problem

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2322    Accepted Submission(s): 940

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

题意：其实X=1，输入n，m ；n行数据，m是模；每行数据输入两个操作，1，x和2，y；意义分别为乘以x、除以第y行数据所对应的x（题目保证第y行为1操作 ）输出每行数据计算后的结果X；

思路：因为题意没有确定m而且m不确定为素数，所以我们不能用逆元来求；利用线段树单点更新（将某个点更新为1），区间（1~n）求积；

代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

typedef long long LL;
const int N=100005;

int n,M;

struct NODE
{
    int l,r;
    LL value;
    int mid(){
        return (l+r)>>1;
    }
}tree[N<<2];

void BuildTree(int l,int r,int rt)
{
    tree[rt].l=l;
    tree[rt].r=r;
    tree[rt].value=1;
    if(l==r)
    {
        tree[rt].value=1;
        return ;
    }

    int m=tree[rt].mid();
    BuildTree(l,m,rt<<1);
    BuildTree(m+1,r,rt<<1|1);
    tree[rt].value=tree[rt<<1].value*tree[rt<<1|1].value%M;
}

void UpdateTree(int l,LL val,int rt)
{
    if(tree[rt].l==l&&tree[rt].r==l)
    {
        tree[rt].value=val;
        return ;
    }
    int m=tree[rt].mid();
    if(l<=m) UpdateTree(l,val,rt<<1);
    if(l>m) UpdateTree(l,val,rt<<1|1);
    tree[rt].value=tree[rt<<1].value*tree[rt<<1|1].value%M;
}

LL Query(int l,int r,int rt)
{
    if(tree[rt].l==l&&tree[rt].r==r)
        return tree[rt].value;

    int m=tree[rt].mid();
    if(r<=m) return Query(l,r,rt<<1);
    else if(l>m) return Query(l,r,rt<<1|1);
    else return (Query(l,m,rt<<1)*Query(m+1,r,rt<<1|1))%M;
}

int main()
{
    int t;
    scanf("%d",&t);
    int cas=0;
    while(t--)
    {
        scanf("%d%d",&n,&M);
        BuildTree(1,n,1);
        printf("Case #%d:\n",++cas);
        for(int i=1;i<=n;i++)
        {
            int a;
            LL b;
            scanf("%d%I64d",&a,&b);

            if(a==1)
                UpdateTree(i,b,1);
            else if(a==2)
                UpdateTree(b,1,1);

            printf("%I64d\n",tree[1].value);
        }
    }
    return 0;
}