C
Uri/beecrowd problem no - 1099 solution in C
#include<stdio.h>
int main(){
int N,X,Y,i,j,sum = 0;
scanf("%d", &N);
for(i = 1; i <= N; i++){
scanf("%d%d", &X, &Y);
if(X > Y){
for(j = Y+1,sum = 0; j < X; j++){
if(j % 2 != 0){
sum += j;
}
}
}else if(X <= Y){
for(j = X+1,sum = 0; j < Y; j++){
if(j % 2 != 0){
sum += j;
}
}
}
printf("%d
", sum);
}
return 0;
}
Uri/beecrowd problem no - 1131 solution in C
#include<stdio.h>
int main(){
int limit = 1,counter = 0,counter1 = 0,counter2 = 0,counter3 = 0;
while(limit != 2){
if(limit == 1){
int score1,score2;
scanf("%d%d", &score1, &score2);
if(score1 > score2){
counter1++;
}else{
counter2++;
}
if(score1 == score2){
counter3++;
}
}
printf("Novo grenal (1-sim 2-nao)
");
counter++;
scanf("%d", &limit);
}
printf("%d grenais
", counter);
printf("Inter:%d
", counter1);
printf("Gremio:%d
", counter2);
printf("Empates:%d
", counter3);
if(counter1 > counter2){
printf("Inter venceu mais
");
}
if(counter1 < counter2){
printf("Gremio venceu mais
");
}
if(counter2 == counter1){
printf("Não houve vencedor
");
}
return 0;
}
Uri/Beecrowd problem no - 1149 solution in C
#include<stdio.h>
int main(){
int A,N,sum = 0,i;
scanf("%d%d", &A, &N);
while(N < 0 || N == 0){
scanf("%d", &N);
}
for(i = 0; i < N; i++){
sum = sum + A + i;
}
printf("%d
", sum);
return 0;
}
Uri/Beecrowd problem no - 1151 solution in C
#include<stdio.h>
int main(){
int N,i,n[100];
scanf("%d", &N);
n[0] = 0;
n[1] = 1;
for(i = 2; i < N; i++){
n[i] = n[i-1] + n[i-2];
}
printf("0");
for(i = 1; i < N; i++){
printf(" %d", n[i]);
}
printf("
");
return 0;
}
BEE/URI problem no 1181 solution in C
#include<stdio.h>
int main(){
double M[12][12],sum = 0.0;
int i,j,L;
char x[2];
scanf("%d", &L);
scanf("%s", x);
for(i = 0; i < 12; i++){
for(j = 0; j < 12; j++){
scanf("%lf", &M[i][j]);
}
}
for(i = 0; i < 12; i++){
sum += M[L][i];
}
if(x[0] == 'S'){
printf("%.1lf
", sum);
}else if(x[0] == 'M'){
printf("%.1lf
", sum/12.0);
}
return 0;
}
Uri/Beecrowd Problem no - 1184 solution in C
#include<stdio.h>
int main(){
double M[12][12],sum = 0;
int i,j,counter = 1;
char O[2];
scanf("%s", O);
for(i = 0; i < 12; i++){
for(j = 0; j < 12; j++){
scanf("%lf", &M[i][j]);
}
}
for(i = counter; i < 12; i++){
for(j = 0; j < counter; j++){
sum += M[i][j];
}
counter++;
}
if(O[0] == 'S'){
printf("%.1lf
", sum);
}else if(O[0] == 'M'){
printf("%.1lf
", sum / 66.0);
}
return 0;
}
Uri/Beecrowd Problem no - 1185 Solution in Java
import java.util.Scanner;
public class Main{
public static void main (String[] args) {
Scanner obj = new Scanner(System.in);
double M[][] = new double[12][12];
double sum = 0.0;
int i,j,counter = 1;
String O = obj.next();
for(i = 0; i < 12; i++){
for(j = 0; j < 12; j++){
M[i][j] = obj.nextDouble();
}
}
for(i = 10; i >= 0; i--){
for(j = 0; j < counter; j++){
sum += M[i][j];
}
counter++;
}
if(O.equals("S")){
System.out.printf("%.1f
", sum);
}else if(O.equals("M")){
System.out.printf("%.1f
", sum/66.0);
}
}
}