| Бодолт илгээх | Бүх бодолтууд | Шилдэг бодолтууд | Жагсаалт руу буцах |
RGB7002 - Гурвалжин |
Өгөгдсөн гурвалжны периметрийг ол.
Input
Гурвалжны талууд бүхэл тоогоор нэг мөрөнд зайгаар тусгаарлагдан өгөгдөнө.
Output
Гурвалжны периметр.
Example
Input:
3 4 5
Output:
12
| Нэмсэн: | Bataa |
| Огноо: | 2011-01-15 |
| 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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2024-10-22 13:13:48
int main() { int n, sum = 0; // Initialize sum to 0 scanf("%d", &n); for (int i = 1; i <= n; i++) { // Declare i in the for loop sum += i; // Add i to sum } printf("%d\n", sum); // Print the sum with a newline for better output format return 0; } 18 |
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2024-10-21 03:21:08
naad neg zuragaa ywulha bolido |
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2024-10-19 07:33:08
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣠⣤⣤⣤⣄⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣴⣿⣿⣿⣿⣿⣿⣿⣷⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢸⣿⣿⣿⣿⣿⣿⣿⣿⣿⣧⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢸⣿⣿⣿⣿⣿⣿⣿⣿⣿⡿⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⢿⣿⣿⣿⣿⣿⣿⣿⡿⠃⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠉⠻⠿⠿⠿⠟⠋⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣴⣶⣿⣿⣶⣄⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢰⣿⣿⣿⣿⣿⣿⣿⣷⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣿⣿⣿⣿⣿⣿⣿⣿⣿⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣼⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⢠⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⣾⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⣸⣿⣿⣿⣿⣿⣿⣿⣿⡟⣿⣿⣿⣿⣧⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣤⣶⣾⣿⣶⣶⣤⡀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⢠⣿⣿⣿⣿⣿⣿⣿⣿⡿⠀⠘⢿⣿⣿⣿⣷⡀⠀⠀⠀⠀⠀⠀⠀⠀⠠⠀⣴⣿⣿⣿⣿⣿⣿⣿⣿⣿⡄⠀ ⠀⠀⠀⠀⠀⠀⠀⣼⣿⣿⣿⣿⣿⣿⣿⣿⠇⠀⠀⠈⠻⣿⣿⣿⣿⣆⠀⠀⠀⢀⠀⠀⠀⠀⢰⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠀ ⠀⠀⠀⠀⠀⠀⠀⣿⣿⣿⣿⣿⣿⣿⣿⡟⠀⣀⣤⣶⣶⣌⠻⣿⣿⣿⣷⡄⠀⠀⠀⠀⠀⠀⣸⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡟⠀ ⠀⠀⠀⠀⠀⠀⠀⠹⣿⣿⣿⣿⣿⣿⣿⠁⣰⣿⣿⣿⣿⣿⣦⣙⢿⣿⣿⣿⠄⠀⠀⠀⠀⠀⣿⣿⣿⣿⣿⣿⣿⣿⣿⡿⠟⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⢿⣿⣿⣿⣿⣿⣿⠀⣿⣿⣿⣿⣿⣿⣿⣿⣦⣹⣟⣫⣼⣿⣿⣶⣿⣿⣿⣿⣿⣿⣯⡉⠉⠉⠁⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⢸⣿⣿⣿⣿⣿⣿⠀⢸⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡇⠀⠀⠀⠐⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠈⣿⣿⣿⣿⣿⣿⡆⠀⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡇⠀⠁⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⣿⣿⣿⣿⣿⣿⡇⠀⢻⣿⣿⣿⣿⣿⡇⠀⠀⠈⠉⠉⢻⣿⣿⣿⣿⣿⣿⣿⣿⣿⠉⠀⠀⠀⠀⠀⠀⠀ ⠀⣠⣴⣶⣶⣶⣶⣶⣶⣾⣿⣿⣿⣿⣿⡇⠀⠸⣿⣿⣿⣿⣿⡇⠀⠀⠀⠀⠀⠀⠹⢿⣿⣿⢿⣿⣿⣿⡿⠀⠀⠀⠀⠀⠀⠀⠀ ⢸⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡇⢰⣶⣿⣿⣿⣿⣿⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢸⣿⣿⣿⣧⣄⣐⣀⣀⣀⣀⣀⡀ ⠸⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡇⢸⣿⣿⣿⣿⣿⣿⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣼⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿ ⠀⠀⠉⠉⠙⠛⠛⠛⠛⠛⠛⠛⠛⠛⠛⠁⠛⠛⠛⠛⠛⠛⠛⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⠉⠉⠉⠉⠉⠉⠉⠉⠉⠉⠉⠁ |
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2024-10-18 02:22:10
⣴⣾⣿⣿⣶⡄⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⢸⣿⣿⣿⣿⣿⣿⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠈⢿⣿⣿⣿⣿⠏⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠊⣉⣩⣀⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⣼⣿⣿⣿⣷⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⢀⣼⣿⣿⣿⣿⣿⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⢀⣾⣿⣿⣿⣿⣿⣿⣷⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⢠⣾⣿⣿⠉⣿⣿⣿⣿⣿⡄⠀⢀⣠⣤⣤⣀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠙⣿⣿⣧⣿⣿⣿⣿⣿⡇⢠⣿⣿⣿⣿⣿⣧⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠈⠻⣿⣿⣿⣿⣿⣿⣷⠸⣿⣿⣿⣿⣿⡿⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠘⠿⢿⣿⣿⣿⣿⡄⠙⠻⠿⠿⠛⠁⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⡟⣩⣝⢿⠀⠀⣠⣶⣶⣦⡀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⣷⡝⣿⣦⣠⣾⣿⣿⣿⣿⣷⡀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⣿⣿⣮⢻⣿⠟⣿⣿⣿⣿⣿⣷⡀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⣿⣿⣿⡇⠀⠀⠻⠿⠻⣿⣿⣿⣿⣦⡀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⢰⣿⣿⣿⠇⠀⠀⠀⠀⠀⠘⣿⣿⣿⣿⣿⡆⠀⠀ ⠀⠀⠀⠀⠀⠀⢸⣿⣿⣿⠀⠀⠀⠀⠀⠀⣠⣾⣿⣿⣿⣿⠇⠀⠀ ⠀⠀⠀⠀⠀⠀⢸⣿⣿⡿⠀⠀⠀⢀⣴⣿⣿⣿⣿⣟⣋⣁⣀⣀⠀ ⠀⠀⠀⠀⠀⠀⠹⣿⣿⠇⠀⠀⠀⠸⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿ |
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2024-10-16 11:09:18
Enkherdene2011.aternos.me Port: 43543 |
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2024-10-15 11:17:55
#include <stdio.h> int main(void) { // your code here int a ,b,c,d,e,f; scanf ("%d %d %d %d",&a,&b,&c,&d); if (a<5){ a=a+0; }else { a=1; } if (b<5){ b=b+0; }else { b=1; } if (c<5){ c=c+0; }else { c=1; } if (d<5){ d=d+0; }else { d=1; } e=a*b*c*d; printf("%d",e); return 0; } |
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2024-10-15 03:53:58
#include <bits/stdc++.h> using namespace std; int a[1000000]; int dp[1000001]; const int MOD = (int)1e9 + 7; int main() { int n, x; cin >> n >> x; for (int i = 0; i < n; i++) { cin >> a[i]; } dp[0] = 1; for (int i = 1; i <= n; i++) { for (int j = 0; j <= x; j++) { if (j - a[i - 1] >= 0) { dp[j] += dp[j - a[i - 1]]; dp[j] %= MOD; } } } cout << dp[x] ; } |
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2024-10-14 11:22:06
khanaaaaaaaaaaaa6y4k.aternos.me:16754 |
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2024-10-14 06:14:50
Code bichdeg naiztai boln shvv IG:ratitulation8 Holbogdooroi |
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2024-10-14 06:13:54
Henbe ci |
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