杭电oj java printf_杭电OJ的输入输出格式题讲解.doc

杭电OJ的输入输出格式题讲解

杭电OJ的输入/输出格式题

输入

一、1000 A + B Problem

题目名称：A + B Problem

链接地址：/showproblem.php?pid=1000

Time Limit: 1 Seconds????Memory Limit:32768K

Time Limit: 2000/1000 MS (Java/Others) ? ?Memory Limit: 65536/32768 K (Java/Others)

Problem Description

Calculate A + B.

Input

Each line will contain two integers A and B. Process to end of file.

Output

For each case, output A + B in one line.

Sample Input

1 1

Sample Output

2

参考答案

#include

int main(void)

{

int a,b;

while(scanf("%d%d",&a,&b)!=EOF)

printf("%d\n",a+b);

return 0;

}

二、1002 A + B Problem II

题目名称：A + B Problem II

链接地址：/showproblem.php?pid=1002

Time Limit: 1 Seconds????Memory Limit:32768K

Time?Limit:?2000/1000?MS?(Java/Others)?Memory?Limit:?65536/32768?K?(Java/Others)

Total?Submission(s):?22627?Accepted?Submission(s):?4035

Problem?Description

I?have?a?very?simple?problem?for?you.?Given?two?integers?A?and?B,?your?job?is?to?calculate?the?Sum?of?A?+?B.

Input

The?first?line?of?the?input?contains?an?integer?T(1<=T<=20)?which?means?the?number?of?test?cases.?Then?T?lines?follow,?each?line?consists?of?two?positive?integers,?A?and?B.?Notice?that?the?integers?are?very?large,?that?means?you?should?not?process?them?by?using?32-bit?integer.?You?may?assume?the?length?of?each?integer?will?not?exceed?1000.

Output

For?each?test?case,?you?should?output?two?lines.?The?first?line?is?"Case?#:",?#?means?the?number?of?the?test?case.?The?second?line?is?the?an?equation?"A?+?B?=?Sum",?Sum?means?the?result?of?A?+?B.?Note?there?are?some?spaces?int?the?equation.?Output?a?blank?line?between?two?test?cases.

Sample?Input

2

1?2

112233445566778899?998877665544332211

Sample?Output

Case?1:

1?+?2?=?3

Case?2:

112233445566778899?+?998877665544332211?=?1111111111111111110

思路：

??。不超过1000位，也就是说用整型数定义是不行的，然后想到的就是用数组。具体解题思路如下：

??1先定义两个字符型数组，保存，然后再定义整型数组，保存两加数的和

??2、下面来求和计算，字符型数组字符型数组，。

?最后就是格式控制好输出。#include

#include

#include

int main()

{

char str1[1005],str2[1005];

int n,co