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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
#483387#6744. Squareucup-team052#AC ✓71ms3920kbC++142.3kb2024-07-18 16:34:562024-07-18 16:34:56

Judging History

你现在查看的是最新测评结果

  • [2024-07-18 16:34:56]
  • 评测
  • 测评结果:AC
  • 用时:71ms
  • 内存:3920kb
  • [2024-07-18 16:34:56]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

int T;
ll s, t;

ll sum(ll l, ll r) {
    return 1ll * (l + r) * (r - l + 1) / 2;
}

ll getdt(ll x) {
    ll dt = floor(sqrtl(2 * x) + 1.5);
    ++dt;
    if (2 * x - dt * dt + 3 * dt >= 2.25) return dt;
    return dt - 1;
}

ll getnxt(ll x) {
    return x + getdt(x);
}

int main() {
    scanf("%lld", &T);
    while (T--) {
        scanf("%lld%lld", &s, &t);
        ll ans = 0;
        while (t > s) {
            int dt = getdt(t);
            int l = 1, r = dt - 1, res = dt;
            while (l <= r) {
                int mid = l + (r - l) / 2;
                ll nt = t - sum(mid, dt - 1);
                if (nt <= 0 || getdt(nt) != mid) l = mid + 1;
                else res = mid, r = mid - 1;
            }
            if (t - sum(res, dt - 1) <= s) {
                int l = res, r = dt - 1, pos = l;
                while (l <= r) {
                    int mid = l + (r - l) / 2;
                    if (t - sum(mid, dt - 1) <= s) pos = mid, l = mid + 1;
                    else r = mid - 1;
                }
                t -= sum(pos, dt - 1);
                ans += (dt - 1 - pos + 1);
                break;
            }
            t -= sum(res, dt - 1); ans += dt - res;
            dt = getdt(t);
            l = 1; r = dt - 2; res = dt - 1;
            while (l <= r) {
                int mid = l + (r - l) / 2;
                ll nt = t - sum(mid, dt - 2);
                if (nt <= 0 || getdt(nt) != mid + 1) l = mid + 1;
                else res = mid, r = mid - 1;
            }
            if (t - sum(res, dt - 2) <= s) {
                int l = res, r = dt - 2, pos = l;
                while (l <= r) {
                    int mid = l + (r - l) / 2;
                    if (t - sum(mid, dt - 2) <= s) pos = mid, l = mid + 1;
                    else r = mid - 1;
                }
                t -= sum(pos, dt - 2);
                ans += (dt - 2 - pos + 1) * 2ll;
                break;
            }
            t -= sum(res, dt - 2); ans += (dt - res - 1) * 2;
            if (t - dt + 2 > 0 && getdt(t - dt + 2) == dt - 2) {
                t -= (dt - 2); ++ans;
            }
        }
        ans += s - t;
        printf("%lld\n", ans);
    }
    return 0;
}

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 3852kb

input:

2
5 1
1 5

output:

4
3

result:

ok 2 number(s): "4 3"

Test #2:

score: 0
Accepted
time: 71ms
memory: 3920kb

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