HDU5475(线段树)

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

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：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=100005;
typedef long long ll;
struct Node{
    ll val;
    int l,r;
}a[MAXN*3];
int n,mod;
void build(int rt,int l,int r)
{
    a[rt].l=l;
    a[rt].r=r;
    a[rt].val=1;
    if(l==r)
    {
        return ;
    }
    int mid=(l+r)>>1;
    build(rt<<1,l,mid);
    build((rt<<1)|1,mid+1,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)>>1;
    if(pos<=mid)
    {
        update(rt<<1,pos,val);
    }
    else
    {
        update((rt<<1)|1,pos,val);
    }
    a[rt].val=(a[rt<<1].val*a[(rt<<1)|1].val)%mod;
}
int main()
{
    int T;
    scanf("%d",&T);
    for(int cas=1;cas<=T;cas++)
    {
        scanf("%d%d",&n,&mod);    
        build(1,1,n);
        printf("Case #%d:\n",cas);
        for(int i=1;i<=n;i++)
        {
            int type,val;
            scanf("%d%d",&type,&val);
            if(type==1)
            {
                update(1,i,val);
            }
            else
            {
                update(1,val,1);
            }
            printf("%lld\n",a[1].val);
        }
    }
    return 0;
}