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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 |
---|---|---|---|---|---|---|---|---|---|
#601055 | #6439. Cloud Retainer's Game | neetman | RE | 0ms | 3788kb | C++14 | 1.2kb | 2024-09-29 20:43:56 | 2024-09-29 20:44:00 |
Judging History
answer
#include<algorithm>
#include<iostream>
#include<cstdio>
#include<map>
using namespace std;
const int N = 2e5 + 5;
int n, m, H;
struct node{
int x, y, op;
}nod[N];
int main(){
ios::sync_with_stdio(0),cin.tie(0);
int T;
cin >> T;
while(T --){
cin >> H;
cin >> n;
for(int i = 1; i <= n; i ++){
cin >> nod[i].x >> nod[i].y;
nod[i].op = 0;
}
cin >> m;
for(int i = 1; i <= m; i ++){
cin >> nod[i + n].x >> nod[i + n].y;
nod[i + n].op = 1;
}
sort(nod + 1, nod + 1 + n + m, [](node x,node y){
return x.x > y.y;
});
map <int, int> f;
for(int i = 1; i <= n + m; i ++){
if(nod[i].op){
f[(nod[i].x + nod[i].y) % (2 * H)] ++;
f[(0ll + 2 * H - nod[i].y + nod[i].x) % (2 * H)] ++;
}
else{
f[(nod[i].x + nod[i].y) % (2 * H)] = f[(0ll + 2 * H - nod[i].y + nod[i].x) % (2 * H)]
= max(f[(nod[i].x + nod[i].y) % (2 * H)], f[(0ll + 2 * H - nod[i].y + nod[i].x) % (2 * H)]);
}
}
cout << f[0] << '\n';
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 0ms
memory: 3788kb
input:
2 4 3 1 1 2 2 6 2 4 3 1 3 3 5 1 7 3 3 1 4 2 3 1 1 6 2 9 1
output:
3 3
result:
ok 2 number(s): "3 3"
Test #2:
score: -100
Runtime Error
input:
5503 10 19 2 4 2 8 8 3 8 4 8 7 2 7 2 6 1 5 3 2 6 4 2 1 4 5 2 5 7 1 4 7 5 7 2 2 8 6 8 1 12 5 1 4 8 5 2 6 1 3 6 1 1 1 7 7 2 5 6 6 8 1 2 3 5 10 5 9 5 10 7 6 6 5 7 1 3 9 6 8 8 8 6 4 2 9 5 4 4 2 10 9 2 3 2 1 7 1 4 3 14 4 6 6 1 2 1 7 6 2 3 4 4 5 3 6 5 1 4 3 4 3 2 6 2 8 6 8 2 6 6 5 2 5 1 3 1 2 3 7 4 5 5 3 ...