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Ehab and the Expected XOR Problem

發布時間:2024/10/5 编程问答 22 豆豆
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https://codeforces.com/contest/1174/problem/D

題解:

C++版本一

題解:樸素

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=300000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; int ans,cnt,flag,temp,sum; int a[N]; bool b[N]; char str; struct node{}; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d%d",&n,&m);b[0]=b[m]=1;int xo=0;flag=1;int add=1;while(flag){flag=0;a[cnt++]=add;add=1;for(int i=1;i<(1<<n);i*=2){if((!b[i^xo])&&(!b[i^m^xo])){b[i^xo]=1;xo^=i;add=i;flag=1;break;}}}cout<<cnt-1<<endl;for(int i=1;i<cnt;i++)printf("%d%c",a[i]," \n"[i==cnt-1]);//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

C++版本二

題解:規律

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=300000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; int ans,cnt,flag,temp,sum; int a[N]; bool b[N]; char str; struct node{}; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){//while(~scanf("%d%d",&n,&m)){scanf("%d%d",&n,&m);int x=m;sum=0;while(x)sum++,x>>=1;sum--;temp=1<<sum;cnt=1<<(m<(1<<n)?n-1:n);cout<<cnt-1<<endl;for(int i=1;i<cnt;i++)printf("%d%c",(((i&-i)<temp)?(i&-i):((i&-i)<<1))," \n"[i==cnt-1]);//}//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

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