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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#201151 | #7333. Dominoes | kjhhjki | AC ✓ | 1ms | 3564kb | C++20 | 821b | 2023-10-05 12:43:16 | 2023-10-05 12:43:16 |
Judging History
answer
#include <bits/stdc++.h>
#define MAXN 100005
#define For(I,A,B) for(int I = (A), endi = (B); I <= endi; ++I)
#define foR(I,A,B) for(int I = (A), endi = (B); I >= endi; --I)
#define ForE(I,A) for(int I = head[A]; I; I = e[I].nxt)
using namespace std;
typedef long long _ll;
typedef unsigned int ui;
int T,n;
void solve()
{
cin >> n;
if(n == 1) cout << "1 1 1 2\n";
if(n == 2) cout << "1 1 1 2\n1 3 1 4\n1 5 1 6\n";
if(n == 3) cout << "2 1 2 2\n2 3 2 4\n3 3 3 4\n1 4 1 5\n2 5 2 6\n3 5 3 6\n";
if(n == 4) cout << "1 1 1 2\n1 3 2 3\n1 4 1 5\n2 1 3 1\n2 2 3 2\n2 4 3 4\n3 3 4 3\n2 5 3 5\n4 5 4 4\n4 1 4 2\n";
if(n > 4) cout << "impossible\n";
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cin >> T;
while(T--) solve();
return 0;
}
详细
Test #1:
score: 100
Accepted
time: 1ms
memory: 3336kb
input:
1 2
output:
1 1 1 2 1 3 1 4 1 5 1 6
result:
ok OK!
Test #2:
score: 0
Accepted
time: 0ms
memory: 3564kb
input:
50 888 171 6 7 8 9 10 1000 1 2 3 4 5 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 71 172 516 669 997
output:
impossible impossible impossible impossible impossible impossible impossible impossible 1 1 1 2 1 1 1 2 1 3 1 4 1 5 1 6 2 1 2 2 2 3 2 4 3 3 3 4 1 4 1 5 2 5 2 6 3 5 3 6 1 1 1 2 1 3 2 3 1 4 1 5 2 1 3 1 2 2 3 2 2 4 3 4 3 3 4 3 2 5 3 5 4 5 4 4 4 1 4 2 impossible impossible impossible impossible impossib...
result:
ok OK!