HDU 6129 Just do it(规律)

Just do it

HDU 6129 (规律) 2017ACM暑期多校联合训练 - Team 7 1010 Just do it

题目链接

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1234 Accepted Submission(s): 724

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≤2×10^5,1≤m≤10^9).
The second line contains n nonnegative integers a1…n(0≤ai≤2^30−1).

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
1
1 3 1

题意：

给定一个数组a，然后要求出数组b，数组b于数组a之间满足这样的规律：
b[i]=a[1]^a[2]^·····^a[i].

像这样的话其实是很简单的，但是我们并不是要求第一次异或得出的数组b，而是要求经过m次异或之后得出的数组b。

分析：
ans[i][j]数组进行异或

你会发现 ans【i】【j】=ans【i-1】【j】^ans【i】【j-1】；
有木有发现杨辉三角形的痕迹。

对于每一项，他的系数就是杨辉三角的值，那么如果当前位子系数为奇数的话，结果就会有贡献（x=x^y^y）.
同样很显然，我们第i行，第j列的答案，其系数为C（i+j-2，j-1）【此时只考虑a的系数】

#include <iostream>
#include <cstdio>
#include<string.h>
using namespace std;
const int N = 2e6+10;
int a[N], b[N];
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        int n, m;
        scanf("%d %d", &n, &m);
        memset(b,0,sizeof(b));
        for(int i=1; i<=n; i++)
        {
            scanf("%d", &a[i]);
        }
        for(int i=1; i<=n; i++)///第m次异或第i项的时候
        {
            int nn=m+i-2,mm=i-1;
            if((nn&mm)==mm)
            {
                for(int j=i; j<=n; j++)  b[j]^=a[j-i+1];
            }
        }
        for(int i=1; i<=n; i++) printf("%d%c",b[i],i==n?'
':' ');
    }
    return 0;
}