HDU 6129 Just do it

Just do it

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)

Total Submission(s): 439    Accepted Submission(s): 245


Problem Description

There is a nonnegative integer sequence a1...n of
length n.
HazelFan wants to do a type of transformation called prefix-XOR, which means a1...n changes
into b1...n,
where bi equals
to the XOR value of a1,...,ai.
He will repeat it for m times,
please tell him the final sequence.

 

Input

The first line contains a positive integer T(1≤T≤5),
denoting the number of test cases.

For each test case:

The first line contains two positive integers n,m(1≤n≤2×105,1≤m≤109).

The second line contains n nonnegative
integers a1...n(0≤ai≤230−1).

 

Output

For each test case:

A single line contains n nonnegative
integers, denoting the final sequence.

 

Sample Input

2
1 1
1
3 3
1 2 3

 

Sample Output

2017 Multi-University Training Contest - Team 7

题意：

已知一个A【i】数组，需要求一个B【i】数组

B【i】 = A【1】 ^ A【2】... ^ A【i】

求经过m次XOR运算后，所得到的B【i】数组

题解：

打个表后 发现 1-2位 2次循环 2-4位 4次一循环

5-8位8次一循环

基本规律是看最后一位数字所在的2^k次方的区间

m次XOR过程可以减少，但还是妥妥的TLE

从二进制来考虑的话，发现会出现4次一循环

a[i] = a[i] ^ a[i - cnt]推导可得

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<time.h>
#define ll long long
using namespace std;
int main() {
int t,n,m;
ll a[200050];
//	clock_t start,end;
//	start = clock();
scanf("%d",&t);
while (t--) {
scanf("%d%d",&n,&m);
for (int i = 1; i <= n; i++)
scanf("%lld",&a[i]);
int cnt = 1;
while ((cnt << 2) <= m)
cnt <<= 2;
while (m > 0) {
while (m >= cnt) {
for (int i = cnt + 1; i <= n; i++)
a[i] = a[i] ^ a[i - cnt];
m -= cnt;
}
cnt >>= 2;
}
printf("%lld",a[1]);
for (int i = 2; i <= n; i++) {
printf(" %lld",a[i]);
}
puts("");
}
//	end = clock();
//	printf("time = %d",end - start);
return 0;
}