SPOJ.com - Problem RGB7002

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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

Submit solution!

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

hide comments

2024-12-13 06:41:30
https://play.typeracer.com?rt=tskq4mu15

2024-12-12 08:51:53
https://play.typeracer.com?rt=gsrebewnd

2024-12-10 08:00:36
#include <stdio.h>

int main() {
int a,b,c,p;
scanf("%d %d %d",&a,&b,&c);
p=a+b+c;
printf("%d",p);

return 0;
}
mai anda

2024-12-10 04:45:53
zl\
z;z;xSLC<lda
f

2024-12-10 02:00:10

Last edit: 2024-12-10 02:04:37

2024-12-10 01:56:41
#include <iostream>
using namespace std;

int main() {

int a,b,c;

int perimeter;
cin>>a>>b>>c;

perimeter=a+b+c;

cout<<perimeter;

return 0;
}

2024-12-10 01:21:29
#include<bits/stdc++.h>

int main() {
int n;
scanf("%d", &n);

for (int i = 1; i <= n; i++) {
if (n % i == 0) {
printf("%d ", i);
}
}

return 0;
}

2024-12-06 08:20:15
⠀⡠⠞⠉⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⠲⡑⢄⠀⠀⠀⠀⠀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⡼⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠙⢮⣣⡀⠀⠀⠀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⠞⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠁⠱⡀⠀⠀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⡞⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢣⠀⠀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⡞⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⡆⠀
⠀⠀⠀⠀⠀⠀⠀⠀⡼⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣀⣠⡤⠤⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢸⣹⠀
⠀⠀⠀⠀⢀⣀⣀⣸⢧⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣠⠞⣩⡤⠶⠶⠶⠦⠤⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠘⡇⡇
⠀⠀⠀⣰⣫⡏⠳⣏⣿⠇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠚⠁⠀⠀⠀⠀⠀⠀⠙⢿⣿⣶⣄⡀⠀⠀⢀⡀⠀⠀⠀⠀⠀⡀⡅⡇
⠀⠀⢰⡇⣾⡇⠀⠙⣟⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣀⣠⣴⣶⠿⠛⠻⢿⣶⣤⣍⡙⢿⣿⣷⣤⣾⡇⣼⣆⣴⣷⣿⣿⡇⡇
⠀⠀⢸⡀⡿⠁⠀⡇⠈⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣸⣿⣿⣯⠴⢲⣶⣶⣶⣾⣿⣿⣿⣷⠹⣿⣿⠟⢰⣿⣿⣿⠿⣿⣿⣿⠁
⠀⠀⠈⡇⢷⣾⣿⡿⢱⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⠉⠉⠹⣌⠳⣼⣿⣿⣿⣻⣿⣿⣿⣿⡇⠈⠁⢰⣿⣿⣿⣿⣶⣾⣿⣿⠀
⠀⠀⠀⣷⠘⠿⣿⡥⠏⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⠃⠌⠉⣿⣿⣿⣿⣿⣿⠟⠃⠀⢀⡿⣿⣿⣿⣿⣿⣿⣿⡞⠀
⠀⠀⠀⢸⡇⠀⠹⠗⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣸⣿⡿⠟⠉⠉⠀⠀⠀⠈⢃⣿⣿⣿⣿⣿⣿⡻⠀⠀
⠀⠀⠀⠈⢧⠀⠀⠏⣇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠙⠋⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⣿⣿⣿⣿⣿⣿⠁⠀⠀
⠀⠀⠀⠀⠈⢳⠶⠞⠃⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣴⠆⠀⠀⠊⠁⠀⠀⠀⠀⠸⣿⣿⣿⣿⣿⣿⠀⠀⠀
⠀⠀⠀⠀⠀⢸⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠂⠀⣼⣿⣀⡰⠀⠀⣤⣄⠀⠀⠀⠀⢹⣿⣿⣿⣿⢻⠀⠀⠀
⠀⠀⠀⠀⠀⡟⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠉⠹⣿⠀⠀⠀⠀⠉⠀⠀⠀⠀⠀⠙⣿⣿⣿⡏⠀⠀⠀
⠀⠀⠀⠀⠀⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢻⣄⢠⣤⣶⣤⣀⠀⢀⣶⣶⣶⣿⣿⠟⠀⠀⠀⠀
⠀⠀⠀⠀⠀⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣠⠖⠁⠀⠀⠀⠀⠀⠻⣿⣿⣥⣤⣯⣥⣾⣿⣿⣿⣿⠋⠀⠀⠀⠀⠀
⠀⠀⠀⠀⣰⠇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⡼⠁⠀⠀⠀⠠⠀⠀⠀⠀⠈⣿⣿⣼⣿⣿⣿⣿⣿⣿⠇⠀⠀⠀⠀⠀⠀
⠀⠀⠀⡰⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠊⠀⠀⠀⣠⣰⣄⡀⠀⢀⣀⣀⣛⣟⣿⣿⣿⣿⣿⣿⡿⠀⠀⠀⠀⠀⠀⠀
⠀⣠⠜⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣠⣼⠾⠛⠛⠻⠿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠃⠀⠀⠀⠀⠀⠀⠀
⠾⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣼⡟⠀⠀⠀⠀⠠⣄⣉⣉⣻⣿⣿⣿⣿⣿⣿⡟⠧⢄⡀⠀⠀⠀⠀⠀⠀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠻⠅⠀⠀⠀⠀⠘⠉⠹⣿⣿⣿⣿⣿⣿⣿⣿⣧⡀⠀⠉⠓⠢⣄⠀⠀⠀
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣉⣿⣿⣿⣿⣿⣿⣿⣿⣷⣻⡄⠀⠀⢀⡑⠢⠄

2024-12-05 02:31:03
#include <bits/stdc++.h>

using namespace std;
int n, m;
char a[1005][1005];

struct Node {
int x, y, cost;
};

char pre[1005][1005];
bool vis[1005][1005];

bool check(int x, int y){
if(x >= 1 && x <= n && y >= 1 && y <= m && a[x][y] != '#' && !vis[x][y]) return true;

return false;
}

int movex[] = {1, -1, 0, 0};
int movey[] = {0, 0, 1, -1};
char movech[] = {'D', 'U', 'R', 'L'};

int main() {

int sx, sy, ex, ey;
cin >> n >> m;
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
cin >> a[i][j];

if(a[i][j] == 'A'){
sx = i;
sy = j;
}

}
}
queue<Node> Q;
Q.push({sx, sy, 0});

string s;

while(!Q.empty()){
Node p = Q.front();
Q.pop();

if(a[p.x][p.y] == 'B'){

cout << "YES" << endl;
cout << p.cost << endl;
for(int i = 0; i < p.cost; i++){
s += pre[p.x][p.y];
// cout << p.x << ' ' << p.y << ' ' << pre[p.x][p.y] << endl;
if(pre[p.x][p.y] == 'U'){
p.x++;
}
else if(pre[p.x][p.y] == 'D'){
p.x--;
}
else if(pre[p.x][p.y] == 'L'){
p.y++;
}
else if(pre[p.x][p.y] == 'R'){
p.y--;
}
}

reverse(s.begin(), s.end());
cout << s;
return 0;
}

for(int i = 0; i < 4; i++){
int x = p.x + movex[i];
int y = p.y + movey[i];
if(check(x, y)){
pre[x][y] = movech[i];
vis[x][y] = true;
Q.push({x, y, p.cost + 1});
}
}
}

cout << "NO";
return 0;
}
gurvaljinii saijruulsan bodolt

2024-12-05 02:30:46
huul2

Last edit: 2024-12-05 02:31:36

АЛГОРИТМЫН ЦАГААН ТОЛГОЙ

RGB7002 - Гурвалжин

Input

Output

Example