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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 |
---|---|---|---|---|---|---|---|---|---|
#630335 | #6744. Square | Qzong# | AC ✓ | 49ms | 3904kb | C++14 | 1.1kb | 2024-10-11 17:59:07 | 2024-10-11 17:59:07 |
Judging History
answer
#include <bits/stdc++.h>
#define int long long
typedef long long s64;
const s64 inf = 1e18;
int go(s64 x) {
return (int)(sqrt(2 * x) + 1.5);
}
s64 calc(s64 st, s64 ed) {
int k = go(st);
int l = 1, r = 10000000000;
while (l < r) {
int mid = (l + r) >> 1;
if ((__int128) (2 * k + mid - 1) * mid / 2 >= ed - st)
r = mid;
else
l = mid + 1;
}
return l;
}
void work() {
s64 x, y;
scanf("%lld%lld", &x, &y);
if (x >= y) {
printf("%lld\n", x - y);
return;
}
int B = calc(x, y);
s64 l = 1, r = x;
while (l < r) {
s64 mid = (l + r) >> 1;
s64 k = go(mid);
if (1ll * (2 * k + B - 1) * B / 2 >= y - mid)
r = mid;
else
l = mid + 1;
}
s64 k = go(l);
s64 ed = l + 1ll * (2 * k + B - 1) * B / 2;
printf("%lld\n", (x - l) + B + (ed - y));
}
signed main() {
// freopen("a.in","r",stdin);
// freopen("a.out","w",stdout);
int T;
scanf("%lld", &T);
while (T --)
work();
return 0;
}
详细
Test #1:
score: 100
Accepted
time: 0ms
memory: 3904kb
input:
2 5 1 1 5
output:
4 3
result:
ok 2 number(s): "4 3"
Test #2:
score: 0
Accepted
time: 49ms
memory: 3816kb
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