Submission #2325717

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```// 三分探索 & 二分探索解法
#include <bits/stdc++.h>
#include <boost/multiprecision/cpp_dec_float.hpp>
#include <boost/multiprecision/cpp_int.hpp>
#define REP(i,n) for (long i=0;i<(n);i++)
#define FOR(i,a,b) for (long i=(a);i<(b);i++)
#define RREP(i,n) for(long i=n;i>=0;i--)
#define RFOR(i,a,b) for(long i=(a);i>(b);i--)
#define dump1d_arr(array) REP(i,array.size()) cerr << #array << "[" << (i) << "] ==> " << (array[i]) << endl;
#define dump2d_arr(array) REP(i,array.size()) REP(j,array[i].size()) cerr << #array << "[" << (i) << "]" << "[" << (j) << "] ==> " << (array[i][j]) << endl;
#define dump(x)  cerr << #x << " => " << (x) << endl;
#define CLR(vec) { REP(i,vec.size()) vec[i] = 0; }
#define llINF (long long) 9223372036854775807
#define loINF (long) 2147483647
#define shINF (short) 32767
#define SORT(c) sort((c).begin(),(c).end())
#define MIN(vec) *min_element(vec.begin(), vec.end());
#define MAX(vec) *max_element(vec.begin(), vec.end());
#define UNIQ(vec) vec.erase(unique(vec.begin(), vec.end()),vec.end());
#define IN(n,m)  (!(m.find(n) == m.end()))
#define TO_INT(vec,s) REP(i,s.length()){vec.push_back(s[i] - ‘0’);}
#define ENUM_v(vec) for (auto e : vec)
#define dump_MAP(m) for(auto itr = m.begin(); itr != m.end(); ++itr) { cerr << itr->first << " --> "  << itr->second << endl; }
#define FINDL(vec,x) (lower_bound(vec.begin(),vec.end(),x) - vec.begin())
#define FINDU(vec,x) (upper_bound(vec.begin(),vec.end(),x) - vec.begin())

using namespace std;
typedef long long LL;
typedef vector<LL> VI;
typedef vector<VI> VVI;
typedef boost::multiprecision::cpp_dec_float_100 dd;
typedef boost::multiprecision::cpp_int ii;

int main(void) {
short Q;
LL A,B,mul,X,tmp,l2,r2,x;
dd l3,r3,q1,q2,range;
cin >> Q;
VI ans(Q);
REP(i,Q){
cin >> A >> B;
if (A > B) {
tmp = A;
A = B;
B = tmp;
}
mul = A*B;
l2 = (A + 1);
r2 = mul;
while (l2 <= r2){
X = (l2+r2)/2;
l3 =  (dd)(A+1);
r3 = (dd)X;
while ((((r3*r3) - (l3*l3)) / (r3+l3)) > 0.01) {
q1 = ((l3*2 + r3)/3);
q2 = ((r3*2 + l3)/3);
if ((q1 / ((dd)X - q2 + 1.0)) >= (q2 / ((dd)X - q1 + 1.0))) r3 = q2;
else l3 = q1;
}
x = (LL) l3;
if (((ii)(X - x + 1)*(ii)x) >= mul) r2 = X-1;
else {
ans[i] = X - 1;
l2 = X+1;
}
}
}

REP(i,Q) cout << ans[i] << endl;
return 0;
}```

#### Submission Info

Submission Time 2018-04-08 19:09:26+0900 D - Worst Case banboooo044 C++14 (GCC 5.4.1) 0 2386 Byte WA 2103 ms 256 KB

#### Test Cases

Set Name Score / Max Score Test Cases
Sample 0 / 0 s1.txt
All 0 / 700 01.txt, 02.txt, 03.txt, 04.txt, 05.txt, 06.txt, 07.txt, 08.txt, 09.txt, 10.txt, 11.txt, s1.txt
Case Name Status Exec Time Memory
01.txt
02.txt
03.txt
04.txt 1055 ms 256 KB
05.txt
06.txt 152 ms 256 KB
07.txt 1746 ms 256 KB
08.txt 1828 ms 256 KB
09.txt 1979 ms 256 KB
10.txt 1819 ms 256 KB
11.txt 3 ms 256 KB
s1.txt 32 ms 256 KB