The Pilots Brothers' refrigerator

The input contains four lines. Each of the four lines contains four characters describing the initial state of appropriate handles. A symbol “+” means that the handle is in closed state, whereas the symbol “−” means “open”. At least one of the handles is initially closed.

The first line of the input contains N – the minimum number of switching. The rest N lines describe switching sequence. Each of the lines contains a row number and a column number of the matrix separated by one or more spaces. If there are several solutions, you may give any one of them.

C++(g++ 7.5.0) 解法, 执行用时: 3ms, 内存消耗: 436K, 提交时间: 2023-07-16 16:09:42

#include<bits/stdc++.h>
using namespace std;
long long ans,t;
vector<pair<long long,long long> >vt;
int main(){
	string s[4];cin>>s[0]>>s[1]>>s[2]>>s[3];
	for(register int i=0;i<4;++i)
		for(register int j=0;j<4;++j){
			t=0;
			for(register int k=0;k<4;++k)if(s[i][k]=='+')++t;
			for(register int k=0;k<4;++k)if(s[k][j]=='+'&&k!=i)++t;
			if(t&1)++ans,vt.push_back(make_pair(i+1,j+1));
		}
	printf("%lld\n",ans);
	for(register int i=0;i<vt.size();++i)printf("%lld %lld\n",vt[i].first,vt[i].second);
}

C++11(clang++ 3.9) 解法, 执行用时: 6ms, 内存消耗: 508K, 提交时间: 2020-01-29 10:33:11

#include<bits/stdc++.h>
using namespace std;
int ans,x[20],y[20];
char a[10][10];bool f[10][10];
int main(){
	for(int i=1;i<5;++i){
		cin>>a[i];
		for(int j=0;j<4;++j){
			if (a[i][j]=='-')f[i][j+1]=0;
			else f[i][j+1]=1;
		}
	}
	for(int i=1;i<5;++i)
		for(int j=1;j<5;++j){
			int t=f[i][j];
			for(int k=1;k<5;++k){
				t^=f[i][k];t^=f[k][j];
			}
			if (t){x[++ans]=i;y[ans]=j;}
		}
	cout<<ans<<endl;
	for(int i=1;i<=ans;++i)cout<<x[i]<<" "<<y[i]<<endl;
	return 0;
}

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