标签:solution break include print stream gauss pre line puts
\(a_{ij}\)以及\(b_i\)都是\(0/1\)
异或运算可以看成是不进位的加法,因此直接高斯消元即可
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 110;
int n;
int a[N][N];
int gauss()
{
int r, c;
for(r = 0, c = 0; c < n; c ++) {
int t = r;
for(int i = r; i < n; i ++) {
if(a[i][c]) {
t = i;
break;
}
}
if(!a[t][c]) continue;
for(int i = c; i < n + 1; i ++) swap(a[r][i], a[t][i]);
for(int i = r + 1; i < n; i ++) {
if(a[i][c]) {
for(int j = n; j >= c; j --) {
a[i][j] ^= a[r][j];
}
}
}
r ++;
}
if(r < n) {
for(int i = r; i < n; i ++) {
if(a[i][n]) {
return 2;
}
}
return 1;
}
for(int i = n - 1; i >= 0; i --) {
for(int j = i + 1; j < n; j ++) {
a[i][n] ^= a[j][n] & a[i][j];
}
}
return 0;
}
int main()
{
scanf("%d", &n);
for(int i = 0; i < n; i ++) {
for(int j = 0; j < n + 1; j ++) {
scanf("%d", &a[i][j]);
}
}
int t = gauss();
if(!t) for(int i = 0; i < n; i ++) printf("%d\n", a[i][n]);
else if(t == 1) puts("Multiple sets of solutions");
else puts("No solution");
return 0;
}
标签:solution break include print stream gauss pre line puts
原文地址:https://www.cnblogs.com/miraclepbc/p/14406279.html