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
--------------------------------------------------------------------------------------------------------------------------
 
//高精度加法,注意格式
#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<memory.h>
#define SIZE 1005 int main()
{
int i, n;
int j, k, m;
int index;
int lengthA, lengthB;
char a[SIZE];
char b[SIZE]; int ans[SIZE];
int carryBit; //进位
//freopen("F:\\input.txt","r",stdin);
scanf("%d", &n);
for(i = 0; i < n; i++)
{
scanf("%s%s", a, b); memset(ans, 0, sizeof(ans));
index = 0;
carryBit = 0;
lengthA = strlen(a);
lengthB = strlen(b); lengthA--;
lengthB--;
while ((lengthA >= 0) && (lengthB >= 0))
{
//将char转换为int,并计算和,如果ans大于10,则carryBit为1,否则为0
ans[index] = (a[lengthA] - '0') + (b[lengthB] - '0') + carryBit;
if (ans[index] >= 10)
{
carryBit = ans[index] / 10;
ans[index] %= 10;
}
else
carryBit = 0;
index++;
lengthA--;
lengthB--;
}
//如果A的长度大于B,则将A与carryBit相加
while (lengthA >= 0)
{
ans[index] = (a[lengthA] - '0') + carryBit;
if (ans[index] >= 10)
{
carryBit = ans[index] / 10;
ans[index] %= 10;
}
else
carryBit = 0;
index++;
lengthA--;
}
//如果B的长度大于A,则将B与carryBit相加
while (lengthB >= 0)
{
ans[index] = (b[lengthB] - '0') + carryBit;
if (ans[index] >= 10)
{
carryBit = ans[index] / 10;
ans[index] %= 10;
}
else
carryBit = 0;
index++;
lengthB--;
} if (carryBit != 0)
{
ans[index]++;
index++;
} //去掉多余的零
while (ans[index] == 0)
index--;
printf("Case %d:\n", i + 1);
printf("%s + %s = ", a, b);
//两个零字符串相加输出为0
if (index == -1)
printf("0");
else
{
for (m = index; m >= 0; m--)
printf("%d", ans[m]);
}
printf("\n");
if (i != n - 1)
printf("\n");
}
//freopen("con", "r", stdin);
//system("pause");
return 0;
}

  

ACM1002:A + B Problem II的更多相关文章

  1. HDU 1002:A + B Problem II(大数相加)

    A + B Problem II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) ...

  2. A + B Problem II

    之前总是在查阅别人的文档,看着其他人的博客,自己心里总有一份冲动,想记录一下自己学习的经历.学习算法有一段时间了,于是想从算法开始自己的博客生涯O(∩_∩)O~~ 今天在网上看了一道大数相加(高精度) ...

  3. hduoj 1002 A + B Problem II

    原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1002 题目描述如下: A + B Problem II Time Limit: 2000/1000 M ...

  4. hdu1032 Train Problem II (卡特兰数)

    题意: 给你一个数n,表示有n辆火车,编号从1到n,入站,问你有多少种出站的可能.    (题于文末) 知识点: ps:百度百科的卡特兰数讲的不错,注意看其参考的博客. 卡特兰数(Catalan):前 ...

  5. nyoj 623 A*B Problem II(矩阵)

    A*B Problem II 时间限制:1000 ms  |  内存限制:65535 KB 难度:1   描述 ACM的C++同学有好多作业要做,最头痛莫过于线性代数了,因为每次做到矩阵相乘的时候,大 ...

  6. HDU 1002 A + B Problem II

    A + B Problem II   Time Limit: 1000MS      Memory Limit: 65536K Total Submissions: 16104    Accepted ...

  7. nyoj 103 A + B problem II

    点击打开链接 A+B Problem II 时间限制:3000 ms  |  内存限制:65535 KB 难度:3 描述 I have a very simple problem for you. G ...

  8. Train Problem II(卡特兰数+大数乘除)

    Train Problem II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) ...

  9. HDU 1002 A + B Problem II(大整数相加)

    A + B Problem II Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u De ...

随机推荐

  1. [翻译] NimbusKit

    注意:NimbusKit 是Github上iOS部分开源库排名前20中的一员. http://nimbuskit.info/ https://github.com/jverkoey/nimbus ht ...

  2. C++实现线性表的链接存储结构(单链表)

    将线性表的抽象数据类型定义在链接存储结构下用C++的类实现,由于线性表的数据元素类型不确定,所以采用模板机制. 头文件linklist.h #pragma once #include <iost ...

  3. 用phpstudy搭建的lnmp环境下mysql授权远程连接

    1.使用phpstudy安装的mysql没有放置到可以直接调用的目录里,所以只能使用绝对路径来访问: /phpstudy/mysql/bin/mysql -uroot -proot 2.执行use m ...

  4. JDK工具系列之jps

    一.简介 jps(JVM Process Status Tool)是虚拟机进程状态工具:可以列出正在运行的虚拟机进程,显示虚拟机正在执行的main()函数,及这些进程的ID(LVMID,Local V ...

  5. Yii 读写分离 分表分库

    本文转自  http://hudeyong926.iteye.com/blog/1299989 实现一主一从,一主多从,多主多从的读写分离 .支持DAO,AR,其中Query builder只完成部分 ...

  6. 「CF375D Tree and Queries」

    题目 \(dsu\ on\ tree\)的板子题了 \(dsu\ on\ tree\)本质上一种优秀通过轻重链剖分优化到\(O(nlogn)\)的暴力 一般用来解决没有修改的允许离线的子树查询问题 首 ...

  7. PHP-------抽象和接口

    静态的关键字是:static Class ren { Public static $yanse;    //yanse是一个静态的成员 Static function show() ; // stat ...

  8. linq 和lamba表达式

    一.什么是Linq(what)二.Linq的优点(why)三.Linq查询的步骤(how)四.查询基本操作五.結合實例代碼(具體聯繫用linqtosql來寫的增刪改查)一.什么是Linq(what). ...

  9. 用@ExceptionHandler 来进行异常处理

    有时候我们想统一处理一个Controller中抛出的异常怎么搞呢? 直接在Controller里面加上用@ExceptionHandler标注一个处理异常的方法像下面这样子 @ExceptionHan ...

  10. js中array(数组).map

    使用前 使用后 代码: