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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
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< Өмнөх 39  40  41  42  43  44  45  46  47  48  49 Дараагийн >
2019-12-01 06:50:33
#include <cstdio>

using namespace std;


int main() {
int n ;
int a[10001],b[10001];
scanf("%d", &n);
for(int i = 1 ; i <= n ; i++){
scanf("%d", &a[i]);
b[i] = a[i];
}
for(int i = 1 ; i < n ; i++){
for(int j = 1 ; j < n ; j++){
if(b[j] > b[j + 1]){
int k = b[j];
b[j] = b[j + 1];
b[j + 1] = k;
}
}
}
for(int i = 1 ; i <= n ; i++){
printf("%d ", a[i] - b[i]);
}
}
2019-12-01 06:11:33
c4c08b3af946023ad6a7
2019-11-30 07:43:29
temuulen63-91916694
tamirta1-2004December27
sorbilig1234-123
2019-11-27 12:04:32
python deerh bodolt bnuu
2019-11-24 07:15:23
5
1 4 2 5 3
#include <cstdio>

using namespace std;


int main() {
int n ;
int a[10001],b[10001];
scanf("%d", &n);
for(int i = 1 ; i <= n ; i++){
scanf("%d", &a[i]);
b[i] = a[i];
}
for(int i = 1 ; i < n ; i++){
for(int j = 1 ; j < n ; j++){
if(b[j] > b[j + 1]){
int k = b[j];
b[j] = b[j + 1];
b[j + 1] = k;
}
}
}
for(int i = 1 ; i <= n ; i++){
printf("%d ", a[i] - b[i]);
}
}
2019-11-24 06:50:20
#include<cstdio>

int main(){
int n ;
scanf("%d", &n);
for(int i = 1; i <= (n +1) / 2 ; i ++){
for(int j = 1; j <= n ; j++)printf("%5d" , (i - 1) * 2 * n + j);
printf("\n");
}
for(int i = (n/2 * 2 -1); i >= 1 ; i -= 2){
for(int j = 1 ; j <= n ; j++)printf("%5d", i * n + j);
printf("\n");
}

}
2019-11-24 06:41:34
ez

Last edit: 2019-11-24 06:42:01
2019-11-24 05:55:07
ez


Last edit: 2019-12-20 10:36:43
2019-11-23 07:32:16
#include <cstdio>
int main() {
long long l,s=0,k=0,a;
scanf("%lld" , &l);
a=(1+l)/2*l;
for(int i=0;i<=l;i=i+7){
s=s+i;
}
k=a-s;
printf("%lld", k);
}


2019-11-23 07:07:25
#include<cstdio>
#include<string>
int main(){
int a,b[255];
scanf("%d", &a);
for(int i=1; i<=a;i++){
scanf("%d", &b[i]);
}
for(int i=1;i<=a;i++){
printf("%c",b[i]);
}
}
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