2015 上海网络赛

An easy problem

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 53    Accepted Submission(s): 22

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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#include 
#include 
#include 
#include 

using namespace std;
#define LL long long
#define lson L, mid, rt << 1
#define rson mid + 1, R, rt << 1 | 1
#define N 100000 + 10

int val[N << 2];
int n, M;

void pushup(int rt)
{
    val[rt] = ((LL)val[rt << 1] * val[rt << 1 | 1]) % M;
}

void build(int L, int R, int rt)
{
    if(L == R)
    {
        val[rt] = 1;
        return ;
    }
    int mid = (L + R) >> 1;
    build(lson);
    build(rson);
    pushup(rt);
}

void update(int q, int v, int L, int R, int rt)
{
    if(L == R)
    {
        val[rt] = v;
        return ;
    }
    int mid = (L + R) >> 1;
    if(q <= mid) update(q, v, lson);
    else update(q, v, rson);
    pushup(rt);
}

int main()
{
    int kase = 0;
    int T;
    scanf("%d", &T);
    while(T--)
    {
        printf("Case #%d:\n", ++kase);
        scanf("%d%d", &n, &M);
        build(1, n, 1);

        int op, y;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d%d", &op, &y);
            if(op == 1)
            {
                update(i, y, 1, n, 1);
                printf("%d\n", val[1]);
            }
            else
            {
                update(y, 1, 1, n, 1);
                printf("%d\n", val[1]);
            }
        }
    }
    return 0;
}


/*

1
10 1000000000
1 2
2 1
1 2
1 10
2 3
2 4
1 6
1 7
1 12
2 7



*/