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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
#744600#9622. 有限小数RosmontispesTL 1ms3616kbC++202.2kb2024-11-13 22:35:432024-11-13 22:35:43

Judging History

你现在查看的是最新测评结果

  • [2024-11-13 22:35:43]
  • 评测
  • 测评结果:TL
  • 用时:1ms
  • 内存:3616kb
  • [2024-11-13 22:35:43]
  • 提交

answer

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

array<int, 3> exgcd(int a, int b, array<int, 4> x = {1, 0, 0, 1}) {
    for (int c; b; tie(a, b) = pair(b, a % b)) {
        c = a / b, x = {x[2], x[3], x[0] - x[2] * c, x[1] - x[3] * c};
    }
    return {x[0], x[1], a};
}

array<int, 3> LiEu(int a, int b, int n = 1, int t = 0) {
    auto [x, y, d] = exgcd(a, b);
    if (n % d) return {x, y, 0};
    t = abs(b) / d, x = ((n / d * x) % t + t) % t;
    return {x, (n - a * x) / b, d};
}

void QAQ() {
    int a, b;
    cin >> a >> b;

    int ta = a, tb = b;
    int d = __gcd(a, b);
    a /= d, b /= d;

    int cnt2 = 0, cnt5 = 0;

    for ( ; ; ) {
        if (ta % 2 && ta % 5) break;
        cnt2 += ta % 2 == 0;
        cnt5 += ta % 5 == 0;
        if (ta % 2 == 0) ta /= 2;
        if (ta % 5 == 0) ta /= 5;
    }

    for ( ; ; ) {
        if (tb % 2 && tb % 5) break;
        cnt2 -= tb % 2 == 0;
        cnt5 -= tb % 5 == 0;
        if (tb % 2 == 0) tb /= 2;
        if (tb % 5 == 0) tb /= 5;
    }

    array<int, 2> ans = {numeric_limits<int>::max(), -1};

    for (int now = 1; now <= 1000000000; now <<= 1) {
        for (int t = now; t <= 1000000000; t *= 5) {
            int d = t * tb;
            int gd = __gcd(d, b);

            auto [c, tt, _] = LiEu(b, tb * b, -a * d);

            if (_ == 0) continue;

            int tmp = (tb * d) / _;

            auto F = [&](int c, int tt, int cnt, int op) {
                for (int i = 0; i <= cnt + 1; i++) {
                    if (c < 0 || tt > 0) continue;
                    int _ = __gcd(c, d);

                    int tc = c / _, td = d / _;
                    if (tc >= 0 && tc < ans[0] && tc <= 1000000000 && td <= 1000000000) {
                        ans = {tc, td};
                    }
                    c -= op * tmp, tt += op * tmp;
                }
            };

            F(c, tt, _, 1), F(c, tt, _, -1);
        }
    }

    cout << ans[0] << " " << ans[1] << "\n";
}

signed main() {
    cin.tie(0) -> sync_with_stdio(0);
    int t = 1;
    cin >> t;

    for (cout << fixed << setprecision(12); t--; QAQ());
}

詳細信息

Test #1:

score: 100
Accepted
time: 1ms
memory: 3616kb

input:

4
1 2
2 3
3 7
19 79

output:

0 1
1 3
1 4375
3 316

result:

ok 4 case(s)

Test #2:

score: -100
Time Limit Exceeded

input:

10000
11 12
28 53
17 60
2 35
17 181
80 123
68 141
79 163
71 99
13 64
33 61
15 32
16 61
11 86
33 74
128 143
40 53
7 23
30 31
5 6
86 181
73 91
13 23
71 81
1 2
7 38
117 160
33 83
129 151
88 153
25 58
16 19
19 141
95 124
43 96
71 139
11 59
106 109
93 152
34 43
17 99
1 57
20 159
16 25
5 73
159 170
172 17...

output:

1 12
1 828125000
1 60
1 7
1 231680000
23 960937500
1 36096000
5 326
1 63360
0 1
1 61000
0 1
1 4880
1 10750
1 18500
1 11714560
1 331250
1 898437500
1 31
1 6
1 289600000
1 455000
1 115000000
1 1265625
0 1
1 1484375
0 1
1 415
1 235937500
1 765000000
1 181250
1 2968750
1 4406250
3 775
1 48
3 347500
1 94...

result: