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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 |
---|---|---|---|---|---|---|---|---|---|
#217166 | #5113. Bridge | ucup-team1001 | RE | 1ms | 5892kb | C++20 | 1.5kb | 2023-10-16 16:09:36 | 2023-10-16 16:09:37 |
Judging History
answer
#include<bits/stdc++.h>
#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
using namespace std;
#define irep(i,l,r) for(int i = l; i <= r; ++ i)
const int N = 100099;
const int len = 153;
const int K = N / len + 50;
int per[N][800],invp[N][800];
stack<array<int, 2>>p[400];
// clo, line
inline int read(){
char ch = getchar();
int s = 0;
while(!isdigit(ch))ch = getchar();
while(isdigit(ch))s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
return s;
}
int main(){
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int n = read(), m = read(), q = read();
for(int c = 0; c <= m / len; c ++){
for(int i = 1; i <= n; ++ i){
per[i][c] = invp[i][c] = i;
}
}
while(q --){
int op = read();
if(op == 1){
int line = read(), cloumn = read();
int x = cloumn / len;
array<int, 2> PI = {cloumn, line};
stack<array<int, 2>>temp;
while(p[x].size()){
auto [c, l] = p[x].top();
if(c < cloumn)break;
temp.push(p[x].top());
per[invp[l][x]][x] = l + 1;
per[invp[l + 1][x]][x] = l;
swap(invp[l][x], invp[l + 1][x]);
p[x].pop();
}
temp.push(PI);
while(temp.size()){
auto [c, l] = temp.top();
per[invp[l][x]][x] = l + 1;
per[invp[l + 1][x]][x] = l;
swap(invp[l][x], invp[l + 1][x]);
p[x].push(temp.top());
temp.pop();
}
}
else{
int a = read();
// cin >> a;
for(int x = 0; x <= m / len; ++ x){
a = per[a][x];
}
cout << a << endl;
}
}
}
詳細信息
Test #1:
score: 100
Accepted
time: 1ms
memory: 5892kb
input:
3 4 13 2 2 1 1 3 2 1 2 2 2 3 1 2 4 2 1 2 2 2 3 1 2 1 2 1 2 2 2 3
output:
2 2 1 3 3 1 2 3 2 1
result:
ok 10 numbers
Test #2:
score: -100
Runtime Error
input:
3 100000 99997 2 2 2 2 2 3 2 3 2 3 2 3 2 3 1 2 11047 1 1 98732 1 2 90045 1 1 43556 2 1 2 3 1 2 17242 1 1 17027 2 1 1 1 94195 2 1 2 2 2 1 2 3 1 1 34124 1 2 14354 1 2 673 1 2 39812 1 2 35520 1 2 16046 2 3 2 2 1 1 25410 2 3 2 1 2 3 2 2 1 2 55684 2 1 1 2 24811 1 2 92268 1 2 60268 2 2 1 1 89272 1 2 19232...
output:
2 2 3 3 3 3 3