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There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import sys input = sys.stdin.readline def solve(): h, c, t = map( int, input().split()) if h <= t: return 1 if 2*t <= h+c: return 2 if 2*h+c <= t*3: if abs(2*h+c-t*3) >= (h-t)*3: return 1 else: return 3 l = 0 r = 500000 whi...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
cases = int(input()) for i in range(cases): l1 = input().split(' ') h = int(l1[0]) c = int(l1[1]) t = int(l1[2]) num = 0 #even number of hot and cold pos1 = (c+h)/2 if t <= pos1: ans = 2 print(ans) else: import math check = math.floor((h-t)/(2*t-h-c))...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import sys input = sys.stdin.readline from collections import * def judge(x): return h*(2*x+1)-x*(h-c)>=(2*x+1)*t def binary_search(): l, r = 0, 10**18 while l<=r: m = (l+r)//2 if judge(m): l = m+1 else: r = m-1 return r T = int(input...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import java.util.Scanner; public class Prob3 { public static void main(String[] args) { Scanner scan = new Scanner(System.in); int T = scan.nextInt(); for (int i = 0; i < T; i++) { int h = scan.nextInt(); int c = scan.nextInt(); int t = scan.nextInt(); ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
""" Python 3 compatibility tools. """ from __future__ import division, print_function import itertools import sys import os from io import BytesIO, IOBase if sys.version_info[0] < 3: input = raw_input range = xrange filter = itertools.ifilter map = itertools.imap zip = itertools.izip def is_it_local(): ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
from fractions import Fraction as frac import sys, os input = sys.stdin.buffer.read().split(b'\n')[::-1].pop from heapq import heappush, heappop def i(): return input() def ii(): return int(input()) def iis(): return map(int, input().split()) def liis(): return list(map(int, input().split())) def pint(x): os.wri...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import java.util.*; public class Main { public static void main(String ars[]) { Scanner sc=new Scanner(System.in); int test=sc.nextInt(); for(int xix=0;xix<test;xix++) { double h=sc.nextInt()+0.0; double c=sc.nextInt()+0.0; double t=sc.nextInt(...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; using ll = long long; double h, c, t; double f(double i) { return (c * (i - 1) + h * (i)) / (2 * i - 1); } void solve() { cin >> h >> c >> t; if (t >= h) { cout << 1 << endl; return; } if (t <= (h + c) / 2) { cout << 2 << endl; return; } ll low =...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
from decimal import Decimal def solve(): h,c,t=[int(x) for x in input().split()] if h==t: print(1) return a=2*t-h-c b=h-t if a<=0: print(2) return else: c1=b//a c2=c1+1 tc1=Decimal((h*(c1+1)+c*c1))/Decimal((2*c1+1)) tc2=Decimal((h*(...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
T = int(input()) for t_itr in range(T): H, C, T = list(map(int, input().rstrip().split())) if T >= H: print(1) elif 2 * T <= H + C: print(2) else: n = (C - T) // (H + C - 2 * T) n1 = n + 1 d1 = H * n + C * (n - 1) - T * (2 * n - 1) d2 = H * n1 + C * (n...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
from decimal import Decimal import sys input=sys.stdin.readline print=sys.stdout.writelines t=int(input()) for i in range(t): h,c,t=map(int,input().split()) avg=(h+c)/2 if t<=avg: k=2 if t>avg: k=int((h-avg)//(t-avg)) if k%2==0: k+=1 ans=k mn=10**7 for ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import java.io.BufferedReader; import java.io.Closeable; import java.io.IOException; import java.io.InputStream; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.StringTokenizer; import static java.lang.Math.min; public class C implements Closeable { private InputReader in = new Inpu...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; int main() { int t; cin >> t; for (int i = 0; i < t; i++) { int h, c, t; cin >> h >> c >> t; if (t >= h) { cout << 1 << endl; } else if (2 * t <= (h + c)) { cout << 2 << endl; } else { double difference = t - (double)(h + c) / 2; ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import math #------------------------------warmup---------------------------- import os import sys from io import BytesIO, IOBase BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in f...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
from decimal import * getcontext().prec=20 def v1(k): num = Decimal((k*h)+((k-1)*c)) den = Decimal((2*k)-1) return num/den def v(k): return ((k*h)+((k-1)*c))/((2*k)-1) import sys # sys.stdin=open("input.txt",'r') # sys.stdout=open("output.txt","w") inf = float('inf') def low(t): l,r = 2,10000000 ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; using namespace std; int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); ; long long int i, j, t, n, k, l, d, r, p, m, x, y; cin >> t; while (t--) { cin >> n >> m >> k; if (k <= (n + m) / 2) { cout << 2 << "\n"; } el...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import sys # from functools import lru_cache, cmp_to_key # from heapq import merge, heapify, heappop, heappush from math import sqrt, ceil, factorial, inf from collections import defaultdict as dd, deque, Counter as C # from itertools import combinations as comb, permutations as perm # from bisect import bisect_left as...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; int main() { int t; cin >> t; while (t--) { long long h, c, t, cnt, ans; cin >> h >> c >> t; if (h + c >= t * 2) { cout << 2 << '\n'; continue; } cnt = (t - c) / (t * 2 - h - c); int x = cnt, y = cnt + 1; double m = (x * h + (x ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
from math import * def r1(t): return t(input().strip()) def r2(t): return [t(i) for i in input().strip().split()] def r3(t): return [t(i) for i in input().strip()] for _ in range(int(input())): h, c, t = r2(int) ans1 = abs((h + c)/2 - t) if ans1 == 0 or 2*t <= (h + c): print(2) e...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools # import time,random,resource # sys.setrecursionlimit(10**6) inf = 10**20 eps = 1.0 / 10**10 mod = 10**9+7 mod2 = 998244353 dd = [(-1,0),(0,1),(1,0),(0,-1)] ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)] def LI(...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; long long x, y, t; void solve() { cin >> x >> y >> t; if (t >= x) { cout << "1\n"; return; } if (t <= (x + y) / 2) { cout << "2\n"; return; } long long ans = (y - t) / (x + y - 2 * t); long long sol = ans; for (int i = max(1LL, ans - 10); i <...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
for i in range (int(input())): a,b,c=map(int,input().split()) d=a-c; e=c-b; f=e-d; if d>=e: print(2) elif d==0: print(1) elif 2<=(e/d): if(c*3-(2*a+b))>=(a-c)*3: print(1) else: print(3) else: g=round((d+e)/(2*(e-d))) ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#!/usr/bin/env python # coding:utf-8 # Copyright (C) dirlt from sys import stdin def run(h, c, t): if (h + c - 2 * t) >= 0: return 2 a = h - t b = 2 * t - h - c k = int(a / b) val1 = abs((k + 1) * h + k * c - (2 * k + 1) * t) val2 = abs((k + 2) * h + (k + 1) * c - (2 * k + 3) * t) ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; void fast() { ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); } long long int gcd(long long int x, long long int y) { if (y == 0) return x; return gcd(y, x % y); } long long int lcm(long long int a, long long int b) { return (a * b) / gcd(a, b);...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
""" This template is made by Satwik_Tiwari. python programmers can use this template :)) . """ #=============================================================================================== #importing some useful libraries. from fractions import Fraction import sys import bisect import heapq from math import...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
q = int(input()) for _ in range(q): h, c, t = map(int, input().split()) if 2 * t <= h + c: print(2) continue l = 1 r = 10 ** 20 while r - l > 1: m = (r + l) // 2 if (m * h + (m - 1) * c >= t * (2 * m - 1)): l = m else: r = m a = l ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
from sys import stdin from collections import Counter, deque from bisect import bisect_left input = stdin.buffer.readline T = int(input()) for _ in range(T): h, c, t = map(int, input().split()) if t*2 <= h+c: print(2); continue a = int((t-h)/(h-2*t+c)) b = a+1 tt = t*(2*a+1)*(2*b+1) aa = abs(tt...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import math def solve(h, c, t): if (h+c)>=2*t: return 2 a = (c-t)/float(h+c-2*t) a1 = math.floor(a) a2 = math.ceil(a) tmp = (2*a2-1)*(a1*h+(a1-1)*c)+(2*a1-1)*(a2*h+(a2-1)*c)-2*t*(2*a1-1)*(2*a2-1) if tmp<=0: return 2*a1-1 else: return 2*a2-1 t = int(input()) for _ in ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import sys from math import inf input = sys.stdin.buffer.readline def prog(): for _ in range(int(input())): h,c,t = map(int,input().split()) lowest = (h+c)/2 if t <= lowest: print(2) elif abs(t-(2*h+c)/3) >= abs(t-h): print(1) else: closest...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import sys input = sys.stdin.buffer.readline for _ in range(int(input())): h, c, t = map(int, input().split()) if h <= t: print(1) elif (h + c) // 2 >= t: print(2) else: n_c = (h - t) // (2 * t - h - c) n_h = n_c + 1 if abs((n_c * c + n_h * h) - t * (n_c + n_h)) ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import math def minOdd (n,m,h,c,t): d = abs((n+1)*(2*m+1)*h + (2*m+1)*n*c - t*(2*n+1)*(2*m+1))-abs((m+1)*(2*n+1)*h + (2*n+1)*m*c - t*(2*n+1)*(2*m+1)) if d>0: return m elif d<0: return n else: return min(n,m) def ans (odd,h,c,t): d = abs(2*(odd+1)*h + 2*odd*c ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
def solver(h, c, t): if h + c >= 2 * t: return 2 elif h == t: return 1 else: x = (t - c) // (2 * t - h - c) k1, k2 = abs((2*x-1)*t- ((h+c)*x-c))*(2*x+1), abs(t*(2*x+1) - ((h+c)*x+h))*(2*x-1) z = 2 * x + 1 if k2 < k1 else 2 * x - 1 return z for _ in ra...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
def solve(h, c, t): if t >= h: return 1 if 2 * t <= h + c: return 2 if (h - t) % (2 * t - h - c) == 0: return 2 * (h - t) // (2 * t - h - c) + 1 nn1 = int((h - t) / (2 * t - h - c)) nn2 = nn1 + 1 if (t * (2 * nn2 + 1) - nn2 * (h + c) - h) * (2 * nn1...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
/** * @author derrick20 */ import java.io.*; import java.util.*; public class MixingWaterDirect { public static void main(String[] args) throws Exception { FastScanner sc = new FastScanner(); PrintWriter out = new PrintWriter(System.out); int T = sc.nextInt(); while (T-->0) { ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; double val(long long x, long long h, long long c) { return ((h * x) + (c * (x - 1))) / ((x * 2.0) - 1.0); } int main() { long long T, h, c, t; cin >> T; while (T--) { cin >> h >> c >> t; if (t <= (h + c) / 2) { cout << 2 << "\n"; continue; } ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import sys from math import ceil, floor from functools import cmp_to_key input = sys.stdin.readline def main(): tc = int(input()) cmp = lambda a, b: a[0] * b[1] - a[1] * b[0] cmp = cmp_to_key(cmp) for _ in range(tc): h, c, t = map(int, input().split()) if h + c == 2 * t: ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
def check(h, c, t, n): return (n + 1) * h + n * c >= (2 * n + 1) * t def binsearch(h, c, t): left = 0 right = 1000000000 while right - left > 1: mid = (right + left) // 2 if check(h, c, t, mid): left = mid else: right = mid if check(h, c, t, right): ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
t=int(input()) for test in range(t): inp=input().split() h=int(inp[0]) c=int(inp[1]) t=int(inp[2]) if t+t <= c+h: print(2) continue if t >= h: print(1) continue k = (h-t)//(t+t-c-h) if (h-t)%(t+t-c-h) == 0: print(k+k+1) continue ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import java.util.*; public class Main { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); public static void main(String[] args) { new Main().run(); } void run() { for (int i = 0; i < n; i++) { long h = sc.nextLong(); long c = sc.nextLong(); ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
for _ in range(int(input())): h,c,t=map(int,input().split()) h-=c t-=c if h==t: print(1) elif h/2>=t: print(2) else: #def f(mid):return h*(1+mid)/(1+2*mid) def p(a): return h*(1+a) def q(a): return 1+2*a a=0 b=10**10...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import java.io.*; import java.util.*; import static java.lang.Math.*; public class Main { public static void main(String[] args) throws IOException { br = new BufferedReader(new InputStreamReader(System.in)); PrintWriter pw = new PrintWriter(System.out); // br = new BufferedReader(new FileRea...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
from math import ceil for _ in range(int(input())): h,c,t=map(int,input().split()) if((h+c)>=t*2): print(2) elif(h-t==0): print(1) else: ans=abs(2*t-c-h) temp=(ceil((h-t)/ans)) res=1000000 hh=0 cc=0 res2=1 for i in range(max(temp,1)...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
def i_by_t(h, c, t): return ((h - c) / (t - (h + c) / 2) + 2) / 2 for _ in range(int(input())): h, c, t = map(int, input().split()) mean = (h + c) / 2 def temp(i): half = i // 2 return (h * (i - i // 2) + c * half) / i T = temp(1) if t >= T: print(1) continue...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> #pragma GCC optimize("Ofast") using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(0); long long T, h, c, t; cin >> T; while (T--) { cin >> h >> c >> t; long double min_dif = fabs((h + c) / 2.0 - t); if (h - t >= t - c) { cout << "2\n"; contin...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; bool sortinrev(const pair<long long, long long> &a, const pair<long long, long long> &b) { return (a.first > b.first); } long long power(long long x, unsigned long long y) { long long p = 998244353; long long res = 1; x = x % p; if (x == 0) return 0...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; const int maxn = 100001; long long h, c, temp; long double f(long long m) { long double r = 1.0 * (m * h + (m)*c) / (2 * m); return (abs(r - temp)); } long double f12(long long m) { long double r = 1.0 * (m * h + (m - 1) * c) / (2 * m - 1); return (abs(r - temp)); }...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import sys from decimal import * input = sys.stdin.readline def main(): t = int(input()) for _ in range(t): H, C, T = [int(x) for x in input().split()] if H == T: print(1) continue if (H + C) / 2 == T: print(2) continue if (H ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; int T; long long h, c, t; int main() { cin >> T; while (T--) { cin >> h >> c >> t; if (h + c >= t * 2) { cout << 2 << '\n'; continue; } long long res = 0; long long n = ceil((h - t) / (2.0 * t - (h + c))); long long res2 = ((n * (h + ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import io,os input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline import sys from math import floor,ceil def solve(h,c,t): if h==t:return 1 if 2*t<=h+c:return 2 if 6*t>=5*h+c:return 1 if 3*t>=2*h+c:return 3 if (h-t)%(2*t-(h+c))==0: cnt=(h-t)//(2*t-(h+c)) return 2*cnt+1 k,l=f...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(0); int t; cin >> t; while (t--) { int h, c, t; cin >> h >> c >> t; if (h + c >= 2 * t) { cout << "2\n"; } else { int k = (h - t) / (2 * t - h - c); if (abs((k + 1) * h + k * c - t...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
from collections import Counter,defaultdict,deque #from heapq import * #import itertools #from operator import itemgetter #from itertools import count, islice #from functools import reduce #alph = 'abcdefghijklmnopqrstuvwxyz' #dirs = [[1,0],[0,1],[-1,0],[0,-1]] #from math import factorial as fact #a,b = [int(x) for x ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
_ = int(input()) for T in range(_): h, c, t = [int(x) for x in input().split()] if t <= (h + c) / 2: print(2) continue x = (t - c) // (2 * t - h - c) v1 = x * h + (x - 1) * c v2 = (x + 1) * h + x * c if abs(v1 - (2 * x - 1) * t) / (2 * x - 1) <= abs(t * (2 * x + 1) - ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import java.util.*; public class solution{ public static void main (String[] args) throws java.lang.Exception { // your code goes here Scanner sc=new Scanner(System.in); int t1=sc.nextInt(); while(t1-->0){ int h=sc.nextInt(); int c=sc.nextInt(); int t=sc.nextInt(); // case 1 n...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; long long h, c, t; double f(long long num) { long long tot = (2 * num - 1); long long val = ceil(tot / 2.) * h + (tot / 2) * c; double avg = ((double)val / tot); double ret = fabs(avg - (double)t); return ret; } void test_cases() { scanf("%lld%lld%lld", &h, &c, ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; int main() { ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL); long long int T; cin >> T; while (T--) { long long int h, c, t; cin >> h >> c >> t; if (t == h) { cout << "1" << "\n"; continue; } if (h + c == 2...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; const int MOD = 1e8; int main() { int q; cin >> q; while (q--) { long long h, c, t, l, r, m; cin >> h >> c >> t; if (t >= h) { cout << "1\n"; continue; } l = 1; r = 1e9; while (r - l > 1) { m = (l + r) / 2; if ((m * ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import sys input = sys.stdin.buffer.readline def I(): return(list(map(int,input().split()))) def sieve(n): a=[1]*n for i in range(2,n): if a[i]: for j in range(i*i,n,i): a[j]=0 return a for __ in range(int(input())): h,c,t=I() # for n in range(1,22,2): # if n%2: # avg=(h+c*(n-1)//2...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; int in(char &c) { return scanf("%c", &c); } int in(char s[]) { return scanf("%s", s); } int in(int &x) { return scanf("%d", &x); } int in(long long &x) { return scanf("%lld", &x); } int in(double &x) { return scanf("%lf", &x); } int in(pair<int, int> &p) { return scanf("%d%...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import sys input = lambda: sys.stdin.readline().rstrip() T = int(input()) for _ in range(T): h, c, t = map(int, input().split()) a = h - t b = t - c if a <= 0: print(1) elif a >= b: print(2) else: mi = (a, 1) mii = 1 if abs((a+b) * mi[1]) < mi[0] * 2: ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; const long long mod = 1000000007; const long long MOD = 998244353; const long long inf = 1e18; const long long MAX = 2e5 + 1; inline long long add(long long a, long long b) { return ((a % mod) + (b % mod)) % mod; } inline long long sub(long long a, long long b) { return...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import java.io.*; import java.util.*; import java.lang.Math; public class Solution { public static void main(String[] args) { Scanner scan = new Scanner(System.in); int T = scan.nextInt(); StringBuilder answer = new StringBuilder(); for (int i = 0; i < T; ++i) { int h = ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import math from decimal import * tt = int(input()) for t in range(tt): x2,x1,x0 = input().split(' ') a1 = int(x1) a2 = int(x2) a0 = int(x0) # if(a0 >= a2): # print(1) # continue if(a0 <= (a1+a2)/2): print(2) continue ans1 = math.floor((a0-a2)/(a1+a2-2*a0)) ans2 = math.ceil((a0-a2)/(a1+a...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import math from sys import stdin,stdout import sys import fractions mod=1000000007 F=fractions.Fraction t=int(stdin.readline()) while t>0: h,c,tt=list(map(int,stdin.readline().split())) mini=sys.maxsize ans=0 if(2*tt>h+c): low=1 high=1000000 mid=low+(high-low)//2 x=0 ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
for _ in range(int(input())): a,b,c=map(int,input().split()) mid=(a+b)/2 if c==a: print(1) elif c<=mid: print(2) elif c>=(mid+(a-b)/6) and c<a: d=(mid+(a-b)/6) if abs(d-c)<abs(a-c): print(3) else: print(1) else: c-=mid ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
t=int(raw_input()) for _ in xrange(t): h,c,t=map(int,raw_input().split()) if 2*t<=h+c: print 2 elif t>=h: print 1 else: k1=(h-t)/(2*t-h-c) f1=[(k1+1)*h+k1*c,2*k1+1] f2=[(k1+2)*h+(k1+1)*c,2*k1+3] s=[f1[0]*f2[1]+f1[1]*f2[0],f1[1]*f2[1]] if s[0]<=2*t*s[1]: print 2*k1+1 else: print 2*k1+3
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import sys from functools import lru_cache, cmp_to_key from heapq import merge, heapify, heappop, heappush from math import * from collections import defaultdict as dd, deque, Counter as C from itertools import combinations as comb, permutations as perm from bisect import bisect_left as bl, bisect_right as br, bisect f...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; const long long INF = 9e18; const long long MOD = 1e9 + 7; const long long mod = 998244353; void fileio() {} int main() { ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); ; fileio(); long long tc; cin >> tc; while (tc--) { long long h, c, t; ci...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#!/usr/bin/env python3 import sys import math from bisect import bisect_right as br from bisect import bisect_left as bl from collections import defaultdict from itertools import accumulate from collections import Counter from collections import deque from operator import itemgetter from itertools import permutations...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import java.io.BufferedInputStream; import java.io.BufferedOutputStream; import java.io.PrintWriter; import java.util.Locale; import java.util.Scanner; public class Main { public static void main(String[] args) throws Exception { try (BufferedInputStream in = new BufferedInputStream(System.in); ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import sys input=sys.stdin.buffer.readline nTests=int(input()) for _ in range(nTests): h,c,t=[int(zz) for zz in input().split()] #Observe that for every nHot==nCold (nCups//2==0), finalt=(h+c)/2. #if t<=(h+c)/2, ans=2 #else, find temp(nHot) just larger than t and find temp(nHot+1) or temp(nHot) is clos...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
ans = [] for _ in range(int(input())): h, c, t = map(int, input().split()) if h + c - 2 * t >= 0: ans.append(2) continue a1 = int((t - c) / (2 * t - h - c)) b1 = a1 - 1 a2 = int((t - c) / (2 * t - h - c)) + 1 b2 = a2 - 1 dt1 = abs(t * (a1 + b1) - (a1 * h + b1 * c)) * (a2 + b2...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
# cook your dish here from sys import stdin, stdout import math from itertools import permutations, combinations from collections import defaultdict from bisect import bisect_left from bisect import bisect_right def L(): return list(map(int, stdin.readline().split())) def In(): return map(int, stdin.readli...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
from sys import stdin from decimal import * inp = lambda: stdin.readline().strip() t = int(inp()) for _ in range(t): h, c, needed = [int(x) for x in inp().split()] s = 10 ** 6 + 1 l = 0 x = 0 prevWell = -1 if needed == h: print(1) elif needed <= (h + c) // 2: print(2) e...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import math import sys input = sys.stdin.readline for _ in range(int(input())): h, c, t = map(int, input().split()) e = 1000001 x = 1 if t > (c+h)/2: a = t - (c+h)/2 x = math.ceil(((h-c)/(2*a) + 1)/2) q = x - 1 if(abs(t - (c+h)/2 - (h-c)/(4*q-2)) <= abs(t - (c+h)/2 - (h...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
T, = map(int, input().split()) def f(h, c, n): return (h*(n+1)+c*n)/(2*n+1) def B(h, c, n): return 2*n+1 def C(h, c, n): return (h*(n+1)+c*n) def H(h, c, t, n): return 2*t*B(h,c,n)*B(h,c,n+1)-C(h,c,n+1)*B(h,c,n)-C(h,c,n)*B(h,c,n+1) >=0 for _ in range(T): h, c, t, = map(int, input().split()) m...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import math from decimal import * for _ in range(int(input())): h,c,t = map(int,input().split()) if t <= (h+c)/2: print(2) else: n = math.ceil((h-c)/((2*t)-(h+c))) # print("before:",n) if n&1: print(n) else: half = n//2 getcontext().prec = 1000 a1 = Decimal((half*h)+((h...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import sys import math from decimal import * t = int(sys.stdin.readline().strip()) for i in range(t): h, c, t = list(map(int,sys.stdin.readline().strip().split(' '))) mid = Decimal(c + (h - c)/2) if h == t: print(1) elif t <= mid: print(2) else: b = int((h - mid)/(t-mid)) ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
for _ in range(int(input())): h,c,t=map(int,input().split()) if(h==t): print(1) else: xx=(h+c)/2 if(xx==t): print(2) elif(xx>t): print(2) else: dif=(t-xx) jj=int(abs((c-h)//(2*dif))) if(jj%2==0): ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
from fractions import Fraction import bisect import os from collections import Counter import bisect from collections import defaultdict import math import random import heapq as hq from math import sqrt import sys from functools import reduce, cmp_to_key from collections import deque import threading from itertools im...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import sys from fractions import Fraction def input(): return sys.stdin.readline().strip() def list2d(a, b, c): return [[c] * b for i in range(a)] def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)] def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in r...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import sys from decimal import Decimal def input(): return sys.stdin.readline().strip() def list2d(a, b, c): return [[c] * b for i in range(a)] def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)] def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in rang...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
for i in range(int(input())): h,c,t=map(int,input().split()) if (h+c)/2>=t: print(2) else: a = (h-t)//(2*t-h- c) b = a+1 print(2*a + 1 if 2*t*(2*a+1)*(2*b+1) >= (2*b+1)*((a+1)*h+a*c)+(2*a+1)*((b+1)*h+b*c) else 2 * b + 1)
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; int main() { int cs; cin >> cs; while (cs--) { double h, c, t; cin >> h >> c >> t; if (t >= h) cout << 1 << endl; else if (t <= (c + h) / 2) cout << 2 << endl; else { long long l = 1, r = 1e9, val; while (l < r) { lo...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> #pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-pointer", "inline") #pragma GCC option("arch=native", "tune=native", "no-zero-upper") #pragma GCC target("avx2") long long read() { long long x = 0, f = 1; char ch = getchar(); while (ch < '0' || ch > '9') { if (ch == '-') f = ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
from math import floor from sys import stdin input = stdin.readline def compare_answer(hot, cold, target, moves, ans): min_temperature = (ans // 2) * (hot + cold) + (ans % 2) * hot min_error = abs(min_temperature - ans * target) * moves temperature = (moves // 2) * (hot + cold) + (moves % 2) * hot err...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
from sys import stdin input=lambda : stdin.readline().strip() char = [chr(i) for i in range(97,123)] CHAR = [chr(i) for i in range(65,91)] mp = lambda:list(map(int,input().split())) INT = lambda:int(input()) rn = lambda:range(INT()) mod = 10000000007 from math import ceil,sqrt,factorial,gcd for _ in rn(): h,c,t = m...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
# cook your dish here import math def bs(H, C, res): lo = 0 hi = 10**12 ans = 0 while(lo<=hi): mid = (lo + hi)//2 if (H*mid + ((mid-1)*C)) >= res*((2*mid) - 1): lo = mid + 1 ans = mid else: hi = mid - 1 return ans def main(): for q i...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
t = int(input()) for test in range(t): h, c, t = map(int,input().split(" ")) t-=c h-=c if t<=h/2: print(2) else: res=int(h/(2*t-h)/2) if(abs(t-(res+1)*h/(2*res+1))>abs(t-(res+2)*h/(2*res+3))): print(2*res+3) else: print(2*res+1) """ So, what h...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
input=__import__('sys').stdin.readline from decimal import Decimal import decimal decimal.getcontext().prec = 20 def cal1(no,h,c): ans = Decimal((h+c)*no + h)/Decimal(2*no+1) return ans for _ in range(int(input())): h,c,t = map(int,input().split()) a1=(h+c)/2 l=1 r=1000000 hh=0 if t<=(h+...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
def getClosest(h, c, st, en, t): if abs(val(h, c, st, t)/(2*st-1)) <= abs(val(h, c, en, t)/(2*en-1)): return st else: return en def val(h, c, x, t): return x * h + (x - 1) * c - (2 * x - 1) * t def findClosest(h, c, st, en, t): # Corner cases if val(h, c, st, t) <= 0: ret...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; const double pi = acos(-1.0); template <typename T, typename U> inline void amin(T &x, U y) { if (y < x) x = y; } template <typename T, typename U> inline void amax(T &x, U y) { if (x < y) x = y; } template <typename T> inline void write(T x) { int i = 20; char buf[...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; double h, c, t; double get(int x) { return fabs(t - (1.0 * x * h + 1.0 * (x - 1) * c) / (1.0 * (x + x - 1))); } int main() { int T; cin >> T; while (T--) { cin >> h >> c >> t; if (h <= t) puts("1"); else { double avg = (h + c) / 2.0; if...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
T=int(input()) for t in range(T): h,c,t=map(int,input().split()) h*=2;c*=2;t*=2 m=(h+c)//2 if t<=m: print(2) else: hc,tc=h-m,t-m tms=hc//tc apr=[2434841,1] for i in range(tms-5,tms+5): if i%2 and abs(tc-hc/i)<apr[0]: apr=[abs(tc-hc/i),i] print(apr[1])
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
#include <bits/stdc++.h> using namespace std; int h, c, t; double ans(double); int binarysearch(); double ghadr(double); bool check(); int main() { int T; cin >> T; while (T--) { cin >> h >> c >> t; if (t <= double(c + h) / 2) { cout << 2 << endl; continue; } if (h == t) { cout <...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import java.io.*; import java.util.*; import static java.lang.Math.*; import static java.util.Arrays.*; public class cf1359c_2 { public static void main(String[] args) throws IOException { int q = ri(); while(q --> 0) { long h = rnl(), c = nl(), t = nl(); if(t <= (h + c) /...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
def solve(h, c, t): if t == h: return 1 if 2 * t <= h + c: return 2 cands = [] cands.append((h + c, 2, 2)) bx = h - t by = 2 * t - h - c b = bx // by for d in range(2): bb = b + d aa = bb + 1 cands.append(( aa * h + bb * c, aa + bb, aa + b...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import sys from math import ceil, floor from functools import cmp_to_key input = sys.stdin.readline def main(): tc = int(input()) for _ in range(tc): h, c, t = map(int, input().split()) if h + c == 2 * t: print(2) else: x = max(0, (t - h) // (h + c - 2 * t)) ...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
from math import ceil for _ in range(int(input())): h,c,t=map(int,input().split()) if t<=(h+c)/2:print(2) else: k=(h-t)//(2*t-h-c) if abs((2*k+3)*t-k*h-2*h-k*c-c)*(2*k+1)<abs((2*k+1)*t-k*h-h-k*c)*(2*k+3):print(2*k+3) else:print(2*k+1)
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import math tt = int(input()) for test in range(tt): h,c,t = map(int,input().split()) if t<=(h+c)/2: print(2) else: k = (h-t)//(2*t-h-c) # if k hot +1 hot and k cold cups if abs((k*(h+c)+h)-t*(2*k+1))*(2*k+3)<=abs(((k+1)*(h+c)+h)-t*(2*k+3))*(2*k+1): print(2*k+1) e...
There are two infinite sources of water: * hot water of temperature h; * cold water of temperature c (c < h). You perform the following procedure of alternating moves: 1. take one cup of the hot water and pour it into an infinitely deep barrel; 2. take one cup of the cold water and pour it into an infin...
import math test = int(input()) for _ in range(test): h,c,t = map(int, input().split()) if t==h: print(1) elif t <= (h+c)//2: print(2) else: x = (t-h)/((h+c)-2*t) ceil = math.ceil(x) floor = math.floor(x) if abs(((h+c)*ceil+h)-t*(2*ceil+1))*(2*floor+1) < abs(((h+c)*floor+...