Educational Codeforces Round 48 D Vasya And The Matrix

EDU #48 D

题意：给定一个矩阵，已知每一行和每一列上数字的异或和，问矩阵上的数字是多少，不存在则输出NO。

思路：构造题，可以考虑只填最后一行，和最后一列，其中（n，m）要特判一下。其他格子给0即可。

　　　自己之前接触这类题目较少，感觉写这种题，自己的智商都提高了。

　　　　

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <list>
#include <cstdlib>
#include <iterator>
#include <cmath>
#include <iomanip>
#include <bitset>
#include <cctype>
using namespace std;
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
#define endl '\n'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //

template<typename T>
inline T read(T&x){
    x=;int f=;char ch=getchar();
    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
    return x=f?-x:x;
}
// #define _DEBUG;         //*//
#ifdef _DEBUG
freopen("input", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
/*-----------------show time----------------*/
                const int maxn = ;
                ll a[maxn],b[maxn];
                ll q[maxn],p[maxn];
                int n,m;
                ll ans[maxn][maxn];
int  main(){
                    OKC;
                    cin>>n>>m;
                    ll s1=,s2 = ;
                    for(int i=; i<=n; i++)cin>>a[i], s1 ^=a[i];
                    for(int j=; j<=m; j++)cin>>b[j], s2 ^=b[j];
                    if(s1!=s2){
                        puts("NO");
                        return ;
                    }
                    ll ss  =s1;
                    cout<<"YES"<<endl;
                    for(int i=; i<=n; i++)
                    {
                        for(int j=; j<=m; j++)
                        {
                                if(i<n&&j<m)cout<<"0 ";
                                else {
                                    if(i==n&&j==m){
                                        cout<<(ss ^ a[n] ^ b[m]) <<" ";
                                    }
                                    else if(i==n){
                                        cout<<b[j]<<" ";
                                    }
                                    else if(j==m){
                                        cout<<a[i]<<" ";
                                    }
                                }
                        }

                        cout<<endl;
                    }

                    return ;
}

EDU#48D