HDU-6356(ST表)

                                                                                       Glad You Came

                                                               Time Limit: 10000/5000 MS (Java/Others)

Problem Description

Steve has an integer array a of length n (1-based). He assigned all the elements as zero at the beginning. After that, he made m operations, each of which is to update an interval of a with some value. You need to figure out ⨁ni=1(i⋅ai) after all his operations are finished, where ⨁ means the bitwise exclusive-OR operator.

In order to avoid huge input data, these operations are encrypted through some particular approach.

There are three unsigned 32-bit integers X,Y and Z which have initial values given by the input. A random number generator function is described as following, where ∧ means the bitwise exclusive-OR operator, << means the bitwise left shift operator and >> means the bitwise right shift operator. Note that function would change the values of X,Y and Z after calling.

Let the i-th result value of calling the above function as fi (i=1,2,⋯,3m). The i-th operation of Steve is to update aj as vi if aj<vi (j=li,li+1,⋯,ri), where

⎧⎩⎨⎪⎪lirivi=min((f3i−2modn)+1,(f3i−1modn)+1)=max((f3i−2modn)+1,(f3i−1modn)+1)=f3imod230(i=1,2,⋯,m).

Input

The first line contains one integer T, indicating the number of test cases.

Each of the following T lines describes a test case and contains five space-separated integers n,m,X,Y and Z.

1≤T≤100, 1≤n≤105, 1≤m≤5⋅106, 0≤X,Y,Z<230.

It is guaranteed that the sum of n in all the test cases does not exceed 106 and the sum of m in all the test cases does not exceed 5⋅107.

Output

For each test case, output the answer in one line.

Sample Input

4

1 10 100 1000 10000

10 100 1000 10000 100000

100 1000 10000 100000 1000000

1000 10000 100000 1000000 10000000

Sample Output

1031463378

1446334207

351511856

47320301347

題意：給你3*m個數的生成規則，用來生成m個區間和對應的值，将區間内小于此值的改為此值.問最後所有值乘對應位置的異或值.

思路：RMQ思想和ST表，可見另一篇文章https://blog.csdn.net/nka_kun/article/details/81461902

然後就是保留區間最值，然後按照區間從大到小枚舉區間，不斷更新其1/2大小的子區間，直到枚舉的區間長度為1.

代碼：

#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
const unsigned int mod = 1<<30;
const int maxn = 2e5+5;
const double eps = 1e-12;
const int inf = 0x3f3f3f3f;
map<int,int>::iterator it;

int n,m;
int lg[maxn];
unsigned int stmax[maxn][18];
unsigned int x,y,z;

unsigned int rng61()
{
    x = x^(x<<11);
    x = x^(x>>4);
    x = x^(x<<5);
    x = x^(x>>14);
    unsigned int w = x^(y^z);
    x = y;
    y = z;
    z = w;
    return z;
}

void init()
{
	lg[0] = -1;
	for(int i = 1;i< maxn;i++)
		lg[i] = ((i&(i-1)) == 0)?(lg[i-1]+1):lg[i-1];
}

int main()
{
	int t;
	cin>>t;
	init();
	
	while(t--)
	{
		scanf("%d %d %u %u %u",&n,&m,&x,&y,&z);
		
		for(int i = 1;i<= m;i++)
		{
			ll f1=rng61();
	        ll f2=rng61();
	        unsigned int f3=rng61();
	        int l=(int)min((f1%n)+1,(f2%n)+1);
	        int r=(int)max((f1%n)+1,(f2%n)+1);
	        unsigned int  v = f3%mod;
			int k = lg[r-l+1];//對應log2的值
			stmax[l][k] = max(stmax[l][k],v);
			stmax[r-(1<<k)+1][k] = max(stmax[r-(1<<k)+1][k],v);
		}
		for(int i = lg[n];i>= 1;i--)
		{
			for(int j = 1;j<= n;j++)
			{
				stmax[j][i-1] = max(stmax[j][i-1],stmax[j][i]);
				stmax[j+(1<<(i-1))][i-1] = max(stmax[j+(1<<(i-1))][i-1],stmax[j][i]);
				stmax[j][i] = 0;
			}
		}
		
		ll ans = 0;
		for(ll i = 1;i<= n;i++)
		{
			ans^= (stmax[i][0]*i);
			stmax[i][0] = 0;
		}
		
		printf("%lld\n",ans);
	}	
	
	return 0;
}