hdu5475 An easy problem（线段树）

思路：一个线段树裸题

#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ULL;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double eps = 1e-9;
const int maxn = 100000 + 100;
int MOD;
struct Node
{
    int l, r;
    LL val;
    Node(){}
    Node(int l, int r):l(l), r(r), val(1){}
}seg[maxn<<2];
void pushup(int x)
{
    seg[x].val = (seg[x<<1].val*seg[x<<1|1].val)%MOD;
}
void Build(int x, int L, int R)
{
    seg[x] = Node(L, R);
    if(L+1==R) return;
    int m = (L+R)>>1;
    Build(x<<1, L, m);
    Build(x<<1|1, m, R);
    pushup(x);
}
void Update(int x, int L, int R, int val)
{
    if(L<=seg[x].l&&seg[x].r<=R)
    {
        seg[x].val = val;
        return;
    }
    int m = (seg[x].l+seg[x].r)>>1;
    if(L<m) Update(x<<1, L, R, val);
    if(m<R) Update(x<<1|1, L, R, val);
    pushup(x);
}


int main()
{
    int T, cas=1, q;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &q, &MOD);
        int x, y;
        Build(1, 1, q+1);
        printf("Case #%d:\n", cas++);
        for(int i=1;i<=q;++i)
        {
            scanf("%d%d", &x, &y);
            if(x==1) Update(1, i, i+1, y);
            else Update(1, y, y+1, 1);
            printf("%lld\n", seg[1].val);
        }
    }
    
    return 0;
}

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