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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#834983#9864. Coinhos_lyricWA 5ms3892kbC++142.1kb2024-12-28 08:39:452024-12-28 08:39:49

Judging History

你现在查看的是最新测评结果

  • [2024-12-28 08:39:49]
  • 评测
  • 测评结果:WA
  • 用时:5ms
  • 内存:3892kb
  • [2024-12-28 08:39:45]
  • 提交

answer

#include <cassert>
#include <cmath>
#include <cstdint>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <functional>
#include <iostream>
#include <limits>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

using namespace std;

using Int = long long;

template <class T1, class T2> ostream &operator<<(ostream &os, const pair<T1, T2> &a) { return os << "(" << a.first << ", " << a.second << ")"; };
template <class T> ostream &operator<<(ostream &os, const vector<T> &as) { const int sz = as.size(); os << "["; for (int i = 0; i < sz; ++i) { if (i >= 256) { os << ", ..."; break; } if (i > 0) { os << ", "; } os << as[i]; } return os << "]"; }
template <class T> void pv(T a, T b) { for (T i = a; i != b; ++i) cerr << *i << " "; cerr << endl; }
template <class T> bool chmin(T &t, const T &f) { if (t > f) { t = f; return true; } return false; }
template <class T> bool chmax(T &t, const T &f) { if (t < f) { t = f; return true; } return false; }
#define COLOR(s) ("\x1b[" s "m")



int main() {
  for (int numCases; ~scanf("%d", &numCases); ) { for (int caseId = 1; caseId <= numCases; ++caseId) {
    Int N, K;
    scanf("%lld%lld", &N, &K);
    
    /*
    for (Int u = 1; ; ) {
      const Int v = u + (u + (K-1) - 1) / (K-1);
      if (v > N) {
        printf("%lld\n", u);
        break;
      }
      u = v;
    }
    */
    Int ans = 0;
    for (Int u = 1; ; ) {
      const Int d = (u + (K-1) - 1) / (K-1);
      /*
        min m s.t.
          ceil((u + d m) / (K-1)) > d
          u/d + m > K-1
        go m steps
      */
      const Int m = max((u + d - 1) / d - K, 1LL);
      if (u + m * d > N) {
        ans = u + ((N - 1) - u) / d * d;
        break;
      }
      u += m * d;
    }
    printf("%lld\n", ans);
  }
#ifndef LOCAL
  break;
#endif
  }
  return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 5ms
memory: 3800kb

input:

4
6 2
8 3
10000 2
1919810 114514

output:

4
8
8192
1919805

result:

ok 4 number(s): "4 8 8192 1919805"

Test #2:

score: -100
Wrong Answer
time: 0ms
memory: 3892kb

input:

100
2 2
2 3
2 4
2 5
2 6
2 7
2 8
2 9
2 10
2 11
3 2
3 3
3 4
3 5
3 6
3 7
3 8
3 9
3 10
3 11
4 2
4 3
4 4
4 5
4 6
4 7
4 8
4 9
4 10
4 11
5 2
5 3
5 4
5 5
5 6
5 7
5 8
5 9
5 10
5 11
6 2
6 3
6 4
6 5
6 6
6 7
6 8
6 9
6 10
6 11
7 2
7 3
7 4
7 5
7 6
7 7
7 8
7 9
7 10
7 11
8 2
8 3
8 4
8 5
8 6
8 7
8 8
8 9
8 10
8 11
9 ...

output:

2
1
1
1
1
1
1
1
1
1
2
3
2
2
2
2
2
2
2
2
4
3
4
3
3
3
3
3
3
3
4
5
4
5
4
4
4
4
4
4
4
5
6
5
6
5
5
5
5
5
4
5
6
7
6
7
6
6
6
6
8
8
8
7
8
7
8
7
7
7
8
8
8
9
8
9
8
9
8
8
8
8
8
9
10
9
10
9
10
9
8
8
11
9
10
11
10
11
10
11

result:

wrong answer 2nd numbers differ - expected: '2', found: '1'