An easy problem

Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1553 Accepted Submission(s): 697

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

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:

504

思路：该题用long long 类型运算会溢出,结果WA。用高精度会TLE。线段树思路：第i个叶子结点维护的是第i个操作.若为乘运算则将结点的值修改为乘数，若为除运算则将结点的值修改为1.父结点维护左右子结点之积。每次输出结果为根节点的值。

#include <cstdio>

#include <algorithm>

using namespace std;

const int MAXN=;

typedef long long ll;

struct Node{

    ll val;

    int l,r;

}a[MAXN*];

int n,mod;

void build(int rt,int l,int r)

{

    a[rt].l=l;

    a[rt].r=r;

    a[rt].val=;

    if(l==r)

    {

        return ;

    }

    int mid=(l+r)>>;

    build(rt<<,l,mid);

    build((rt<<)|,mid+,r);

}

void update(int rt,int pos,int val)

{

    if(a[rt].l==pos&&a[rt].r==pos)

    {

        a[rt].val=val%mod;

        return ;

    }

    int mid=(a[rt].l+a[rt].r)>>;

    if(pos<=mid)

    {

        update(rt<<,pos,val);

    }

    else

    {

        update((rt<<)|,pos,val);

    }

    a[rt].val=(a[rt<<].val*a[(rt<<)|].val)%mod;

}

int main()

{

    int T;

    scanf("%d",&T);

    for(int cas=;cas<=T;cas++)

    {

        scanf("%d%d",&n,&mod);

        build(,,n);

        printf("Case #%d:\n",cas);

        for(int i=;i<=n;i++)

        {

            int type,val;

            scanf("%d%d",&type,&val);

            if(type==)

            {

                update(,i,val);

            }

            else

            {

                update(,val,);

            }

            printf("%lld\n",a[].val);

        }

    }

    return ;

}

巴特西

HDU5475(线段树)

An easy problem

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