2017多校联合第七场1010/hdu 6129 Just do it（递推／杨辉三角）

Just do it

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1203    Accepted Submission(s): 702

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

 

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题目大意：

设定b【i】=a【1】^a【2】^a【3】^..................a【i】；

每进行一次，我们可以从a数组得到一个b数组。问进行m次的结果。

思路：

我们随手写下四项的前两次结果，不难看出，我们设定ans【i】【j】表示进行到第i次，第j个位子的答案的话，ans【i】【j】有推导式：

ans【i】【j】=ans【i-1】【j】^ans【i】【j-1】；

那么很显然，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。

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】；

那么我们只需要考虑第一项（a）对所有位子的结果的影响即可（因为b就相当于向后挪了一下递推即可）。

那么根据上述公式，考虑第一项（a）对所有位子的结果的影响然后递推一下就行了。

#include <iostream>#include <algorithm>#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <string.h>#include <map>#include <set>#include <queue>#include <deque>#include <list>#include <bitset>#include <stack>#include <stdlib.h>#define lowbit(x) (x&-x)#define e exp(1.0)const int mod=1e9+7;const int inf=0x3f3f3f;//ios::sync_with_stdio(false);//    auto start = clock();//    cout << (clock() - start) / (double)CLOCKS_PER_SEC;typedef long long ll;typedef long long LL;using namespace std;const int maxn=2e5+10;ll a[maxn];ll b[maxn];int main(){    ios::sync_with_stdio(false);    int T;    cin>>T;    while(T--)    {        memset(b,0,sizeof(b));        int n,m;        cin>>n>>m;        for(int i=1;i<=n;i++)            cin>>a[i];        for(int i=1;i<=n;i++)        {            int x=i+m-2;            int y=i-1;            if((x&y)==y)            {                for(int j=i;j<=n;j++)                    b[j]^=a[j-i+1];            }        }        for(int i=1;i<n;i++)            cout<<b[i]<<' ';        cout<<b[n]<<endl;    }    return 0;}