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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 |
---|---|---|---|---|---|---|---|---|---|
#630717 | #6744. Square | Calculatelove | AC ✓ | 56ms | 3940kb | C++14 | 911b | 2024-10-11 20:08:01 | 2024-10-11 20:08:01 |
Judging History
answer
#include <bits/stdc++.h>
typedef long long s64;
s64 go(s64 x) {
return (s64)(sqrt(2 * x) + 1.5);
}
__int128 calc(s64 k, s64 b) {
return (__int128) (2 * k + b - 1) * b / 2;
}
s64 calc_b(s64 x, s64 y) {
s64 k = go(x);
s64 l = 1, r = 10000000000;
while (l < r) {
s64 mid = (l + r) >> 1;
if (calc(k, mid) >= y - x) r = mid; else l = mid + 1;
}
return l;
}
s64 calc_a(s64 x, s64 y, s64 b) {
s64 l = 1, r = x;
while (l < r) {
s64 mid = (l + r) >> 1;
s64 k = go(mid);
if (calc(k, b) >= y - mid) 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;
}
s64 b = calc_b(x, y);
s64 a = calc_a(x, y, b);
printf("%lld\n", (x - a) + b + (a + calc(go(a), b) - y));
}
int main() {
int T;
scanf("%d", &T);
while (T --)
work();
return 0;
}
详细
Test #1:
score: 100
Accepted
time: 0ms
memory: 3880kb
input:
2 5 1 1 5
output:
4 3
result:
ok 2 number(s): "4 3"
Test #2:
score: 0
Accepted
time: 56ms
memory: 3940kb
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