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Friday, March 25, 2016

• URI Online Judge Solution | 1004 Simple Product Using C, C++ and Java language

Simple Product

Solution in c language:, solution in c++ language:.

Solution in java language:

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URI Online Judge Solution : 1001 Extremely Basic (Beginner Problem)

Problem summary:.

Problem Number: 1001 Problem Name: Extremely Basic Author’s Name: Adapted by Neilor Tonin, URI Brazil Timelimit: 1 Problem Category: BEGINNER Problem Source: https://www.urionlinejudge.com.br/judge/en/problems/view/1001

Some Talks about Contest Programming:

An incredible method to enhance your abilities when figuring out how to code is by solving coding problems. Solving different kinds of challenges and riddles can enable you to improve as a problem solver, take in the complexities of a programming dialect, get ready for prospective job interviews, learn new algorithms and more.

An online judge is an online platform to test programs in focused programming challenges. They are likewise used to practice for such challenges. A considerable amount of these platforms also arrange their own programming contests.

10 Steps to Solve Any Problems:

• Read the problem completely at least two or three times (or however many makes you feel comfortable)
• Identify the subject, the problem belongs to. Is it a sorting or pattern matching problem? Can I use graph theory? Is it related to number theory? etc.
• Try to solve the problem manually by considering 3 or 4 sample data sets.
• After concentrate on optimizing the manual steps. Try to make it as simple as possible.
• Write to write pseudo-code and comments besides the code from the manual steps. One thing you can do is to check after every function is written. Use a good IDE with a debugger, if possible. Don’t need to think much about the syntax. Just focus on the logic and steps.
• Replace the comments or pseudo-code with real code. Always check if the values and code are behaving as expected before moving to the new line of pseudo-code.
• Then optimize the real code.
• Take care of boundary conditions as well.
• Get feedback from your teammates, professors, and other developers and also ask your question on Stack Overflow if possible. Try to learn from others’ guidelines and what they are handling those problems. A problem may be solved in several ways. So, don’t get disappointed if you can’t think like an expert. You need to stick to the problem and you will gradually become better and quicker in solving problems like others.
• Practice, Practice, and Practice.

N.B: Try to follow the above steps always. If you still can’t get the problem solved, take a look at the solution below. Don’t just copy paste the code. It will kill your creativity. Try to enjoy contest programming and develop your skills.

#include <iostream> using namespace std; int main(){ int a, b; cin >> a; cin >> b; cout << "X = " << a+b << endl; return 0; } N.B.: Code is Collected from Different Sources

Extremely Basic

  Solution:     #include<stdio.h> int main() { int A, B, X; scanf("%d %d", &A, &B); X = A + B; printf("X = %dn", X); return 0; }  

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Do You Feel Your Code?!?!?

Before seeing the solution please make sure that you tried enough., sunday 27 march 2016, solution of uri 1145 :: logical sequence 2.

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18 comments:

if(b==X) printf("%d",a); else printf("%d ", a); if(b==X) { b=0; printf("\n"); } Can I use same conditions ? Like u use if(a==x) twice.

man your code is readless

I has been tried to solve this problem . But always seems the presentation error . #include int main() { int i,n,a,b=0; scanf("%d %d",&a,&n); for(i=1; i<=n; i++) { printf("%d ", i); b++; if(b==a) { printf("\n"); b=0; } } return 0; } l whats the problem

WHY presentation error !

#include int main() { int X,Y,j,i; scanf("%d%d", &X,&Y); if(Y>X) for(i=1;i<=Y;i=i+X) { for(j=i;j<X+i;j++) { printf("%d ", j); } printf("\n"); } return 0; } PRESENTATION ERROR WHY? please,,,,

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#include int main() { int x,y,i,j=0; scanf("%d %d",&x,&y); for(i=1;i<=y;i++) { printf("%d",i); j++; if(j<x) printf(" "); else { printf("\n"); j=0; } } return 0; }

#include using namespace std; main(){ int x,y,i,j,a; cin >> x; cin >> y; if((x>1 && x<20) && (y>x) && y<100000){ for(i=1;i<=y;i++){ a=i; for(j=1;j<=x;j++){ if(a==y){ cout << a << " "; a++; break; } else{ cout << a << " " ; a++; } } cout << endl; i+=x-1; } } } why presentation error

#include void main(void){ int x, y, cont; scanf("%i", &x); do{ scanf("%i", &y); }while(y <= x); for (cont = 1;cont <= y; cont++){ if (cont % x == 0 || cont == y){ printf("%i\n", cont); } else{ printf("%i ", cont); } } }

#include int main () { int n,a,h; scanf ("%d%d",&a,&h); for (n=1;n<=h;n++) { if (n%a==0) { printf ("%d \n",n); } else { printf ("%d ",n);} } return 0; }

What is the problem in this code can anyone help me #include int main() { int i, x, y, j, c = 1; scanf("%d %d", &x, &y); for (i = 1; i <= y; i++) { for (j = 1; j <= x; j++) { if (c == y + 1) { break; } if (c == x) { printf("%d", c++); } else { printf("%d ", c++); } } printf("\n"); if (c == y + 1) { break; } } }

# include int main () { int x,y; scanf("%d%d", &x,&y); int i; for (i=1; i<=y; i++) { if((i%x)==0) { printf("%d\n", i); } else { printf("%d ",i); } } return 0; }

#include #include int main() { int i, s, j, k, temp=0; scanf("%d %d", &s, &j); for(i=1; i<=j; i++) { printf("%d ",i); temp+=1; if(temp%s==0) { printf("\n"); } } return 0; }

why I'm getting presentation error?

#include int main() { int x, y, i, j, n, m; scanf("%d%d",&x,&y); for(i=1;i<=y;i++) { if(i%x==0) { printf("%d\n",i); } else { printf("%d ",i); } } return 0; }

very simple

#include using namespace std; int main() { int x, y; cin >> x >> y; for (int i = 1; i <= y; i++) { cout << i << " "; if (i % x == 0) cout << endl; } } i got presention error

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URI - BEECROWD - BEE 1018 | Banknotes Solution in C,C++,Python

Uri - beecrowd - bee  online judge solution  1018 | banknotes - uri - beecrowd - bee 1018 solution in c,c++,python    .

In this problem you have to read an integer value and calculate the smallest possible number of banknotes in which the value may be decomposed. The possible banknotes are 100, 50, 20, 10, 5, 2 e 1. Print the read value and the list of banknotes.

The input file contains an integer value  N  (0 <  N  < 1000000).

Print the read number and the minimum quantity of each necessary banknotes in Portuguese language, as the given example. Do not forget to print the end of line after each line, otherwise you will receive  “Presentation Error” .

URI Online Judge Solution  1018 | Banknotes - URI 1018 Solution in C,C++,Python::

Uri online judge solution  1017 | fuel spent- uri 1017 solution in c,c++,python , uri online judge solution  1019 | time conversions- uri 1019 solution in c,c++,python , 5 responses to uri - beecrowd - bee 1018 | banknotes solution in c,c++,python.

Thanks Nice Thinking

can anyone tell me why this code is not accepted ? #include int main() { int notes[] = {100,50,20,10,5,2,1}; int A,B; //bool C = true; scanf("%d",&A); B = A; while(B!=0 && 0<A && A<1000000){ for(int i = 0; i < 7; i++){ printf("%d nota(s) de R$ %d,00\n",(B/notes[i]),notes[i]); B = B%notes[i]; } } return 0; }

Good Solution

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Resolução do exercício 1018 disponível no site Uri Online Judge, utilizando a linguagem C#

deborahmancilha/exercicio-uri-1018

Folders and files, repository files navigation, uri 1018 - cédulas.

"Leia um valor inteiro. A seguir, calcule o menor número de notas possíveis (cédulas) no qual o valor pode ser decomposto. As notas consideradas são de 100, 50, 20, 10, 5, 2 e 1. A seguir mostre o valor lido e a relação de notas necessárias."

"O arquivo de entrada contém um valor inteiro N (0 < N < 1000000)."

"Imprima o valor lido e, em seguida, a quantidade mínima de notas de cada tipo necessárias, conforme o exemplo fornecido. Não esqueça de imprimir o fim de linha após cada linha, caso contrário seu programa apresentará a mensagem: “Presentation Error”."

Exemplo entrada:

Exemplo saída:.

5 nota(s) de R$ 100,00

1 nota(s) de R$ 50,00

1 nota(s) de R$ 20,00

0 nota(s) de R$ 10,00

1 nota(s) de R$ 5,00

0 nota(s) de R$ 2,00

1 nota(s) de R$ 1,00"