HDU 6129 Just do it【杨辉三角】【思维题】【好题】

Just do it

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 635    Accepted Submission(s): 356

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

21 113 31 2 3

 

Sample Output

11 3 1

 

Source

2017 Multi-University Training Contest - Team 7

判断每位数对后面的影响即可，

打表发现其每位对后面的值为：

对比杨辉三角，则：

则可直接根据杨辉三角的公式退出每位对后面的影响，则问题解决。

#include <bits/stdc++.h>#define INF 0x3f3f3f3f#define ms(x,y) memset(x,y,sizeof(x))using namespace std;typedef long long ll;const int mod = 1e9 + 7;const int maxn = 2e5 + 100;int a[maxn], b[maxn];int main(){//freopen("in.txt", "r", stdin);//freopen("out.txt", "w", stdout);int t;scanf("%d", &t);while (t--){int n, m;ms(b, 0);scanf("%d%d", &n, &m);for (int i = 1; i <= n; i++)scanf("%d", &a[i]);for (int i = 1; i <= n; i++)//第i位{int x = m + i - 2;//组合数x取yint y = i - 1;if ((x & y) == y)//x取y奇偶判断,如果为奇则对后面有影响{for (int j = 1; j <= n; j++)//计算对后面影响{if (j - i + 1 >= 1)b[j] ^= a[j - i + 1];}}}for (int i = 1; i <= n; i++){printf("%d", b[i]);if (i != n) printf(" ");}puts("");}return 0;}