https://ac.nowcoder.com/acm/contest/327/E

C++版本一

std

題解：

#include <bits/stdc++.h> using namespace std; int n,m; int a[2000000]; int b[2000000];int ones(int x) {int ans=0;while (x){ans++;x=x&(x-1);}return ans; }int main() {//freopen(".\\data\\1.in","r",stdin);//freopen(".\\data\\1.out","w",stdout);scanf("%d%d",&n,&m);if (m%2==0){printf("NO\n");return 0;}a[0]=0;a[1]=1;for (int i=1;i<n;i++){for (int j=(1<<i);j<(1<<(i+1));j++){a[j]=(1<<i)^(a[(1<<(i+1))-1-j]);}}if (m==1){printf("YES\n");for (int i=0;i<(1<<n);i++){printf("%d\n",a[i]);}}else{printf("YES\n");for (int i=0;i<(1<<(m+1));i++){if (i%2==0) b[i]=a[i];else b[i]=a[i]^((1<<(m+1))-1);}for (int i=m+1;i<n;i++){int p=-1;for (int j=0;j<(1<<i);j++){if (ones(b[j]^b[(1<<i)-1])==m-1 && ones(b[(j+(1<<i)-1)%(1<<i)])==m-1){p=j;break;}}assert(p!=-1);for (int j=0;j<(1<<i);j++){b[(1<<i)+j]=(1<<i)^b[(p+j)%(1<<i)];}}for (int i=0;i<(1<<n);i++){printf("%d\n",b[i]);}}return 0; }掃一掃，把題目裝進口袋

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總結

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