| Бодолт илгээх | Бүх бодолтууд | Шилдэг бодолтууд | Жагсаалт руу буцах |
RGB7019 - Илэрхийлэл 1 |
Өгөгдсөн илэрхийллийн утгыг ол. A=B*C-D
Input
Нэг мөрөнд 3 бүхэл тоо нэг нэг зайгаар тусгаарлагдан өгөгдөнө.
Output
Илэрхийллийн нэг мөрөнд гарна.
Example
Input: 2 3 4 Output: 2
| Нэмсэн: | Bataa |
| Огноо: | 2013-01-01 |
| Хугацааны хязгаарлалт: | 1s |
| Эх кодын хэмжээний хязгаарлалт: | 50000B |
| Memory limit: | 1536MB |
| Cluster: | Cube (Intel G860) |
| Програмчлалын хэлүүд: | ADA95 ASM32 BASH BF C NCSHARP CSHARP C++ 4.3.2 CPP C99 CLPS LISP sbcl LISP clisp D ERL FORTRAN HASK ICON ICK JAVA JS-RHINO JULIA LUA NEM NICE OCAML PAS-GPC PAS-FPC PERL PHP PIKE PRLG-swi PYTHON PYPY3 PYTHON3 RUBY SCALA SCM guile ST TCL WHITESPACE |
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2024-05-06 04:25:37
⣴⣾⣿⣿⣶⡄⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⢸⣿⣿⣿⣿⣿⣿⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠈⢿⣿⣿⣿⣿⠏⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠊⣉⣩⣀⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⣼⣿⣿⣿⣷⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⢀⣼⣿⣿⣿⣿⣿⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⢀⣾⣿⣿⣿⣿⣿⣿⣷⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⢠⣾⣿⣿⠉⣿⣿⣿⣿⣿⡄⠀⢀⣠⣤⣤⣀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠙⣿⣿⣧⣿⣿⣿⣿⣿⡇⢠⣿⣿⣿⣿⣿⣧⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠈⠻⣿⣿⣿⣿⣿⣿⣷⠸⣿⣿⣿⣿⣿⡿⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠘⠿⢿⣿⣿⣿⣿⡄⠙⠻⠿⠿⠛⠁⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⡟⣩⣝⢿⠀⠀⣠⣶⣶⣦⡀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⣷⡝⣿⣦⣠⣾⣿⣿⣿⣿⣷⡀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⣿⣿⣮⢻⣿⠟⣿⣿⣿⣿⣿⣷⡀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⣿⣿⣿⡇⠀⠀⠻⠿⠻⣿⣿⣿⣿⣦⡀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⢰⣿⣿⣿⠇⠀⠀⠀⠀⠀⠘⣿⣿⣿⣿⣿⡆⠀⠀ ⠀⠀⠀⠀⠀⠀⢸⣿⣿⣿⠀⠀⠀⠀⠀⠀⣠⣾⣿⣿⣿⣿⠇⠀⠀ ⠀⠀⠀⠀⠀⠀⢸⣿⣿⡿⠀⠀⠀⢀⣴⣿⣿⣿⣿⣟⣋⣁⣀⣀⠀ ⠀⠀⠀⠀⠀⠀⠹⣿⣿⠇⠀⠀⠀⠸⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠇ |
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2024-04-22 03:28:33
no huulah no huulah no huulah no huulah no huulah no huulah no huulah no huulah no huulah no huulah no huulah no huulahno huulah no huulah no huulah no huulah no huulahno huulah no huulah no huulah no huulah no huulah no huulahno huulahno huulahno huulah no huulah no huulah no huulah no huulah no huulah no huulahno huulahno huulah no huulahno huulah no huulah no huulah no huulahno huulah no huulah no huulah no huulahno huulah no huulahno huulahno huulah no huulahno huulahno huulah no huulahno huulahno huulah no huulah no huulah no huulahno huulah no huulah no huulah no huulah no huulah |
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2024-04-20 08:44:42
⠀⢀⣠⣴⣶⣿⣿⣿⣷⣦⡤⠴⠶⠟⠛⠛⠛⠿⠿⢶⣦⣄⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⣠⣴⣾⣿⣿⣿⣿⣿⣿⠿⠛⠛⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⠛⢿⣦⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⢀⣾⣿⣿⣿⣿⣿⣿⣿⠏⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⣻⣶⣶⣶⣤⡀⠀⠀⠀⠀⠀gege u are this dog ⠀⠀⠘⣿⣿⣿⣿⣿⣿⡿⠁⠀⠀⠀⢀⣠⣄⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⠻⣿⣿⣿⣿⣦⡀⠀⠀⠀ ⠀⠀⠀⠹⣿⣿⣿⣿⣿⠃⠀⠀⠀⢠⣿⣿⣀⣿⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠘⣿⣿⣿⣿⣿⣆⠀⠀ ⠀⠀⠀⠀⣿⣿⣿⣿⠃⠀⠀⠀⠀⠀⠙⠿⠟⠃⠀⠀⠀⠀⠀⠀⠀⢰⡏⠙⣿⡆⠀⠀⠀⢸⣿⣿⣿⣿⣿⣧⡀ ⠀⠀⠀⠀⢸⣿⣿⠃⠀⠀⠀⠀⠀⠀⠀⠀⠀⣤⣶⣶⣶⣦⣄⡀⠀⠘⢿⣿⡿⠇⠀⠀⠀⢸⣿⣿⣿⣿⣿⣿⣷ ⠀⠀⠀⠀⠈⠉⠁⠀⠀⠀⠀⠀⡆⠀⠀⠀⠀⢻⣹⣿⣿⣿⡟⣿⡆⠀⠀⠀⠀⠀⠀⠀⠀⢸⣿⣿⣿⣿⣿⣿⡟ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣷⠀⠀⠀⠀⠈⠻⣿⣿⣿⠷⠋⠀⠀⠀⠀⠀⠀⠀⠀⠀⢸⣿⣿⣿⣿⠟⠁⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢸⣷⣄⣀⣀⣀⣴⣿⣧⠀⠀⠀⠀⠀⠀⣠⠃⠀⠀⠀⠀⢸⣿⣿⡿⠃⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢸⣿⣿⣿⠿⠟⠋⠈⢿⣷⣤⣀⣀⣠⡶⠃⠀⠀⠀⠀⠀⠀⠻⠟⠁⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⣿⣿⣏⠀⠀⠀⠀⠀⠉⣿⣿⣿⡿⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠸⣿⣿⠀⠀⠀⠀⠀⣼⣿⣿⠏⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠹⣿⣷⣄⣀⣤⣾⣿⠟⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⣀⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⠛⠻⠿⠛⠋⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⢠⣤⡀⣾⣿⣿⠀⣤⣤⡄⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⢿⣿⡇⠘⠛⠁⢸⣿⣿⠃⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠈⣉⣤⣾⣿⣿⡆⠉⣴⣶⣶⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣀⣄⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⣾⣿⣿⣿⣿⣿⣿⡀⠻⠟⠃⠀⠀⠀⠀⠀⠀⠀⢀⣤⣄⢰⣿⣿⠇⣠⣴⣦⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠙⠛⠻⢿⣿⣿⣿⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠸⣿⣿⠀⠛⡋⠀⣿⣿⡟⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠈⠙⠋⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢈⣡⣴⣿⣿⣷⠀⢠⣶⣶⡆⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢰⣿⣿⣿⣿⣿⣿⣇⠘⠛⠛⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⠛⠻⠿⢿⣿⣿⣿⠄⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠉⠋⠁⠀⠀⠀ |
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2024-04-15 07:05:46
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣠⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⢀⣤⠔⠂⠀⠀⠀⢤⣤⣤⣤⣄⣀⣀⣀⣀⡀⠀⢀⣀⣀⣀⡀⠀⠀⠀⢀⢀⣀⣠⣤⣴⣶⣾⠉⠉⠙⣶⠂⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠰⣿⣷⣾⣿⣷⣴⣶⣾⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠏⠉⠛⢿⣿⣿⣿⣿⣿⣿⣿⠿⠿⣿⣶⣿⣿⣿⣿⣤⣂⣀⡀⠀⠀⠀⠀ ⠀⠀⠀⣿⣿⣿⠿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣧⣤⣶⣶⣿⣿⣿⣿⣿⣿⣿⣯⣤⣤⣿⡿⠿⠿⠿⠛⠛⠛⠛⠋⠀⠀⠀⠀ ⠀⠀⠀⣹⣟⣿⣦⠈⠻⠋⢉⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠏⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⢀⣿⣿⣿⣿⣦⠀⠘⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠿⠟⢿⣿⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⢸⣿⣿⣿⣿⣿⡀⢀⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣦⡈⠋⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠈⠿⠿⠁⣿⣿⣿⠎⠻⣯⠉⠉⠉⠈⠉⠉⠉⠉⠉⠉⠉⠉⠉⠉⢹⣿⢻⠉⠉⠁⠈⠻⢦⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⢹⣿⡟⠀⠀⠈⠳⣄⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⢿⠞⠀⠀⠀⠀⠀⠀⠙⠳⠤⠤⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⢈⡟⠇⠀⠀⠀⠀⠈⠁⠤⠤⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢸⣶⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠙⠀⠈⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⠻⠂ |
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2024-04-15 03:58:28
ds |
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2024-04-08 03:39:02
Last edit: 2024-04-08 03:39:41 |
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2024-03-07 10:54:35
#include<bits/stdc++.h> using namespace std; #define ll long long #define pb push_back #define mk make_pair #define S second #define F first ll t, n; pair < ll, ll > p[200045]; bool check( ll k ){ ll cl = 0, cr = 0; for(int i = 1; i <= n; i ++ ){ cl = max( cl-k, 0LL ); cr = cr+k; if( p[i].S < cl || cr < p[i].F ){ return false; } if( cl <= p[i].F && p[i].S <= cr ){ cl = p[i].F, cr = p[i].S; continue; } if( p[i].F <= cl && p[i].S <= cr ){ cr = p[i].S; continue; } if( cl <= p[i].F && cr <= p[i].S ){ cl = p[i].F; continue; } if( cl <= p[i].F && p[i].S <= cr ) continue; } return true; } int main(){ ios_base::sync_with_stdio(NULL); cin.tie(NULL); cout.tie(NULL); cin >> t; string s; while( t -- ){ cin >> n; for(int i = 1; i <= n; i ++ ) cin >> p[i].F >> p[i].S; ll l = 0, r = 1e9; while( r-l >= 10 ){ ll mid = ( l+r ) / 2; if( check( mid ) == 1 ){ r = mid; }else{ l = mid; } } for(int i = l; i <= r; i ++ ){ if( check(i) == 1 ){ cout << i << "\n"; break; } } } } |
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2024-01-17 08:15:59
Izoxme was here |
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2023-12-05 06:36:25
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2023-11-13 08:34:02
. Last edit: 2023-11-16 02:17:59 |
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