結果

問題 No.1384 Bishop and Rook
ユーザー chocorusk
提出日時 2021-02-10 00:14:33
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,271 ms / 2,000 ms
コード長 2,818 bytes
コンパイル時間 1,607 ms
コンパイル使用メモリ 129,596 KB
最終ジャッジ日時 2025-01-18 16:50:23
ジャッジサーバーID
(参考情報)
judge3 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 55
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <complex>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <cassert>
#include <fstream>
#include <utility>
#include <functional>
#include <stack>
#define popcount __builtin_popcount
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;
int main()
{
int t; cin>>t;
while(t--){
int n, m; cin>>n>>m;
if(n==1 && m==1){
cout<<0<<endl;
cout<<1<<" "<<1<<endl;
continue;
}
if(n%2!=0 || m%2!=0){
cout<<-1<<endl;
continue;
}
cout<<n*m-1<<endl;
auto solve=[&](auto solve, int i)->void{
if(n==2*i || m==2*i) return;
if(i&1){
for(int j=0; j<(m-2*i)/2; j++){
cout<<n-i<<" "<<m-1-i-2*j<<endl;
cout<<n+1-i<<" "<<m-2-i-2*j<<endl;
cout<<n+1-i<<" "<<m-1-i-2*j<<endl;
cout<<n-i<<" "<<m-2-i-2*j<<endl;
}
for(int j=0; j<(n-2-2*i)/2; j++){
if(j==(n-2-2*i)/2-1){
cout<<n-1-i-2*j<<" "<<i<<endl;
cout<<n-2-i-2*j<<" "<<i+1<<endl;
cout<<n-2-i-2*j<<" "<<i<<endl;
cout<<n-1-i-2*j<<" "<<i+1<<endl;
}else{
cout<<n-1-i-2*j<<" "<<i<<endl;
cout<<n-2-i-2*j<<" "<<i+1<<endl;
cout<<n-1-i-2*j<<" "<<i+1<<endl;
cout<<n-2-i-2*j<<" "<<i<<endl;
}
}
}else{
for(int j=0; j<(m-2*i)/2; j++){
cout<<i+2<<" "<<i+1+2*j<<endl;
cout<<i+1<<" "<<i+2+2*j<<endl;
cout<<i+1<<" "<<i+1+2*j<<endl;
cout<<i+2<<" "<<i+2+2*j<<endl;
}
for(int j=0; j<(n-2-2*i)/2; j++){
if(j==(n-2-2*i)/2-1){
cout<<i+3+2*j<<" "<<m-i<<endl;
cout<<i+4+2*j<<" "<<m-i-1<<endl;
cout<<i+4+2*j<<" "<<m-i<<endl;
cout<<i+3+2*j<<" "<<m-i-1<<endl;
}else{
cout<<i+3+2*j<<" "<<m-i<<endl;
cout<<i+4+2*j<<" "<<m-i-1<<endl;
cout<<i+3+2*j<<" "<<m-i-1<<endl;
cout<<i+4+2*j<<" "<<m-i<<endl;
}
}
}
solve(solve, i+1);
};
solve(solve, 0);
}
return 0;
}
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