QOJ.ac
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 |
---|---|---|---|---|---|---|---|---|---|
#387736 | #6744. Square | k1nsom | AC ✓ | 96ms | 3772kb | C++17 | 1.4kb | 2024-04-12 18:58:55 | 2024-04-12 18:58:55 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define N 200005
#define endl '\n'
int T, x, y, n, ret = 0, ans = 0, cnt = 0;
signed main() // int((2x)^(1/2)+1.5)=2,3,3,4,4,4......
{
cin >> T;
while (T--)
{
cin >> x >> y;
// x = rand(), y = rand();
// cout << x << " " << y << endl;
if (x >= y)
{
cout << x - y << endl;
}
else
{
int tmp1 = sqrt(2 * x), tmp2 = sqrt(2 * y);
if (2 * x > tmp1 * (tmp1 + 1))
tmp1++;
tmp1++;
if (2 * y > tmp2 * (tmp2 + 1))
tmp2++;
tmp2++;
cnt = 0;
ret = 0;
ret = x - (tmp1 - 2) * (tmp1 - 1) / 2 + (tmp2 - 1) * tmp2 / 2 - y + tmp2 - tmp1 + 1;
if (tmp1 != tmp2)
{
cnt = tmp2 - tmp1;
x += (tmp2 - 1 + tmp1) * (tmp2 - tmp1) / 2;
tmp1 = tmp2;
}
if (x >= y)
ans = x - y;
else
ans = min(x + tmp2 - y, x - (tmp1 - 2) * (tmp1 - 1) / 2 + (tmp1 - 1) * tmp1 / 2 - y + 1ll);
cout << min(ret, cnt + ans) << endl;
// cout << cnt << ' ' << ans << ' ' << ret << endl;
// cout << tmp1 << ' ' << tmp2 << endl;
}
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 0ms
memory: 3772kb
input:
2 5 1 1 5
output:
4 3
result:
ok 2 number(s): "4 3"
Test #2:
score: 0
Accepted
time: 96ms
memory: 3584kb
input:
100000 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 1 10 1 11 1 12 1 13 1 14 1 15 1 16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 1 26 1 27 1 28 1 29 1 30 1 31 1 32 1 33 1 34 1 35 1 36 1 37 1 38 1 39 1 40 1 41 1 42 1 43 1 44 1 45 1 46 1 47 1 48 1 49 1 50 1 51 1 52 1 53 1 54 1 55 1 56 1 57 1 58 1 59 1 60 1 ...
output:
0 2 1 4 3 2 6 5 4 3 8 7 6 5 4 10 9 8 7 6 5 12 11 10 9 8 7 6 14 13 12 11 10 9 8 7 16 15 14 13 12 11 10 9 8 18 17 16 15 14 13 12 11 10 9 20 19 18 17 16 15 14 13 12 11 10 22 21 20 19 18 17 16 15 14 13 12 11 24 23 22 21 20 19 18 17 16 15 14 13 12 26 25 24 23 22 21 20 19 18 1 0 2 2 1 3 4 3 2 4 6 5 4 3 5 ...
result:
ok 100000 numbers