hdu 5475 （线段树）

An easy problem

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2770    Accepted Submission(s): 1034

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

思路：

之前以为可以直接逆元，后面想起来逆元是要两个互质的数，后面想到了题目就是提示你用线段树，只要用建个线段树就好了。

假如第i次操作  ,x = 1;是第一种操作，那么只要根据修改第x个叶子结点将他修改为y。

x=2，只要将第y个叶子节点修改为1,

每一次操作都求出根节点的值就好了。

实现代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mid int m = (l + r) >> 1
const int M = 1e5+10;
ll sum[M<<2];
ll m;
void pushup(int rt){
    sum[rt] = (sum[rt<<1]*sum[rt<<1|1])%m;
}

void update(int p,int c,int l,int r,int rt){
    if(l == r){
        sum[rt] = c;
        return ;
    }
    mid;
    if(p <= m) update(p,c,lson);
    if(p > m) update(p,c,rson);
    pushup(rt);
}

void build(int l,int r,int rt){
     if(l == r){
        sum[rt] = 1;
        return ;
     }
     mid;
     build(lson);
     build(rson);
     pushup(rt);
}

ll query(int L,int R,int l,int r,int rt){
    if(L <= l&&R >= r){
        return sum[rt];
    }
    mid;
    ll ret = 1;
    if(L <= m) ret= (ret*query(L,R,lson))%m;
    if(R > m) ret= (ret*query(L,R,rson))%m;
    return ret;
}
int main()
{
    ll t,n,x;
    ll y;
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    while(cin>>t){
    int t1 = t;
    while(t--){
        ll ans = 1;
        cin>>n>>m;
        build(1,n,1);
        //cout<<2<<endl;
        cout<<"Case #"<<t1-t<<":"<<endl;
        for(int i = 1;i <= n;i ++){
            cin>>x>>y;
            if(x==1){
                update(i,y,1,n,1);
                cout<<query(1,n,1,n,1)<<endl;
            }
            else{
                update(y,1,1,n,1);
                cout<<query(1,n,1,n,1)<<endl;
            }
        }
        memset(sum,0,sizeof(sum));
    }
    }
    return 0;
}