[acm-icpc] 1 - basic

9
Basic 郭至軒 (KuoE0) [email protected]

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Page 2: [ACM-ICPC] 1 - Basic

Online Judge

• UVa - uva.onlinejudge.org

• POJ - poj.org

• ZeroJudge - zerojudge.tw

• SPOJ - www.spoj.pl

• USACO - ace.delos.com/usacogate

Page 4: [ACM-ICPC] 1 - Basic

Input•多為連續測資(不只⼀一筆)•若未指定測資數量,則讀取至 EOF (end

of file)while ( scanf( “%d”, &x ) != EOF ) {

//do something}

while ( cin >> x ) {//do something

}

Page 5: [ACM-ICPC] 1 - Basic

Output

•「⼀一定」要按照題目規定輸出•多⼀一個空格、字母、換行等等多餘字元,都算錯!

Page 6: [ACM-ICPC] 1 - Basic

Example[POJ] 1000 - A+B Problem

#include <iostream>#include <cstdio>#include <cstdlib>using namespace std;int main() { int a, b; while ( ~scanf( "%d %d", &a, &b ) ) printf( "%d\n", a + b ); return 0;}

Page 7: [ACM-ICPC] 1 - Basic

注意•於 C++ 中使用 scanf/printf 等 function,請引入 cstdio (or stdio.h)

•於 C++ 中使用 atoi/rand 等 function,請引入 cstdlib (or stdlib.h)

•善用好的編輯器(vim, notepad++, Code Block)

Page 8: [ACM-ICPC] 1 - Basic

Practice[POJ] 1207 - The 3n + 1 problem

Page 9: [ACM-ICPC] 1 - Basic

Thank You for Your Listening.