HDU 5475：An easy problem 这题也能用线段树做？？？

An easy problem

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 467    Accepted Submission(s): 258

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

题意是一台计算器，最开始的数字是1，然后对其不断操作，输入的操作为1时，乘以后面的数y。输入的操作为2时，除以第y次操作的数。问每一次操作后的结果是多少。

哇，这题居然是线段树做法，真的是没想到啊，发现线段树的题竟然这么广。

对每一个定点进行更新，对整个线段树求乘积。

代码：

#include <iostream>  
#include <algorithm>  
#include <cmath>  
#include <vector>  
#include <string>  
#include <cstring>  
#pragma warning(disable:4996)  
using namespace std;

struct no
{
	int L,R;
	long long mul;
}tree[400015];

int root,n;
long long mod;

void buildtree(int root,int L,int R)
{
	tree[root].L=L;
	tree[root].R=R;

	tree[root].mul=1;

	if(L!=R)
	{
		buildtree(root*2+1,L,(L+R)/2);
		buildtree(root*2+2,(L+R)/2+1,R);
	}
}

void insert(int root,int s,int e,long long val)
{
	if(tree[root].L==s&&tree[root].R==e)
	{
		tree[root].mul=val%mod;
		return;
	}
	if(e<=(tree[root].L+tree[root].R)/2)
	{
		insert(root*2+1,s,e,val);
	}
	else if(s>=(tree[root].L+tree[root].R)/2+1)
	{
		insert(root*2+2,s,e,val);
	}
	else
	{
		insert(root*2+1,s,(tree[root].L+tree[root].R)/2,val);
		insert(root*2+2,(tree[root].L+tree[root].R)/2+1,e,val);
	}
	tree[root].mul = (tree[2*root+1].mul *  tree[2*root+2].mul)%mod;
}

int main()
{
	//freopen("i.txt","r",stdin);
	//freopen("o.txt","w",stdout);

	int test,i,j,oper;
	long long val;
	scanf("%d",&test);
	
	for(i=1;i<=test;i++)
	{
		printf("Case #%d:\n",i);
		scanf("%d%lld",&n,&mod);

		buildtree(0,1,n);
		for(j=1;j<=n;j++)
		{
			scanf("%d%lld",&oper,&val);
			if(oper==1)
			{
				insert(0,j,j,val);
			}
			else
			{
				insert(0,val,val,1);
			}
			printf("%lld\n",tree[0].mul);
		}
	}
	//system("pause");
	return 0;
}

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