东莞网站建设方案咨询,深圳优化网站公司,一级域名做网站,.net网站开发框架CF1157G. Inverse of Rows and Columns
Solution
首先枚举第一行是否变换#xff0c;再枚举第一行的010101状态#xff0c;即可确定列变换。 然后对于之后的行变换#xff0c;从前往后贪心地让111出现得尽可能晚即可。
Code
#include vector
#include list再枚举第一行的010101状态即可确定列变换。 然后对于之后的行变换从前往后贪心地让111出现得尽可能晚即可。
Code
#include vector
#include list
#include map
#include set
#include deque
#include queue
#include stack
#include bitset
#include algorithm
#include functional
#include numeric
#include utility
#include sstream
#include iostream
#include iomanip
#include cstdio
#include cmath
#include cstdlib
#include cctype
#include string
#include cstring
#include ctime
#include cassert
#include string.h
//#include unordered_set
//#include unordered_map
//#include bits/stdc.h#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i(a);i(b);i)
#define fi first
#define se secondusing namespace std;templatetypename Tinline bool upmin(T x,T y) { return yx?xy,1:0; }
templatetypename Tinline bool upmax(T x,T y) { return xy?xy,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pairint,int PR;
typedef vectorint VI;const lod eps1e-11;
const lod piacos(-1);
const int oo130;
const ll loo1ll62;
const int mods1e97;
const int MAXN205;
const int INF0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f1,x0; char cgetchar();while (c0||c9) { if (c-) f-1; cgetchar(); }while (c0c9) { x(x3)(x1)(c^48); cgetchar(); }return x*f;
}
int a[MAXN][MAXN],X[MAXN],Y[MAXN],Flag[MAXN];
signed main()
{int nread(),mread();for (int i1;in;i)for (int j1;jm;j) a[i][j]read();for (int i0;im;i){int flag1;for (int j1;ji;j) Y[j]a[1][j]^X[1];for (int ji1;jm;j) Y[j]a[1][j]^X[1]^1;for (int j1;jn;j) Flag[j]X[j]0;for (int j2,lst(i!m);jn;j){for (int k1;km;k)if (!lst){if (Flag[j](a[j][k]^X[j]^Y[k])) lst1;if (!Flag[j]k1(a[j][k]^X[j]^Y[k])) lst1;if (!Flag[j]k1(a[j][k]^X[j]^Y[k])) X[j]Flag[j]1;}else{if (Flag[j]!(a[j][k]^X[j]^Y[k])) flag0;if (!Flag[j](a[j][k]^X[j]^Y[k])) X[j]0,Flag[j]1;if (!Flag[j]!(a[j][k]^X[j]^Y[k])) X[j]1,Flag[j]1;}}if (flag){puts(YES);for (int j1;jn;j) putchar(X[j]0);puts();for (int j1;jm;j) putchar(Y[j]0);puts();return 0;} }X[1]1;for (int i0;im;i){int flag1;for (int j1;ji;j) Y[j]a[1][j]^X[1];for (int ji1;jm;j) Y[j]a[1][j]^X[1]^1;for (int j1;jn;j) Flag[j]X[j]0;for (int j2,lst(i!m);jn;j){for (int k1;km;k)if (!lst){if (Flag[j](a[j][k]^X[j]^Y[k])) lst1;if (!Flag[j]k1(a[j][k]^X[j]^Y[k])) lst1;if (!Flag[j]k1(a[j][k]^X[j]^Y[k])) X[j]Flag[j]1;}else{if (Flag[j]!(a[j][k]^X[j]^Y[k])) flag0;if (!Flag[j](a[j][k]^X[j]^Y[k])) X[j]0,Flag[j]1;if (!Flag[j]!(a[j][k]^X[j]^Y[k])) X[j]1,Flag[j]1;}}if (flag){puts(YES);for (int j1;jn;j) putchar(X[j]0);puts();for (int j1;jm;j) putchar(Y[j]0);puts();return 0;} }puts(NO);return 0;
}
