杭电1002(简单的大数)
2015-10-17 16:17
495 查看
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
由于涉及数据的长度约为1000,因此用数组保存数据。由低到高一位一位相加,注意进位,尤其是最高位的进位。此外,本题的格式注意在输出A + B = Sum中间是有空格的,最后一个实例输出后只有一个换行,没有空行。其余任意两个实例之间有空行。
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
由于涉及数据的长度约为1000,因此用数组保存数据。由低到高一位一位相加,注意进位,尤其是最高位的进位。此外,本题的格式注意在输出A + B = Sum中间是有空格的,最后一个实例输出后只有一个换行,没有空行。其余任意两个实例之间有空行。
#include<stdio.h> #include<string.h> char a[1010],b[1010]; int c[1010],d[1010]; int main() { int T,len1,len2,len,i,j,k; scanf("%d",&T); for(k=1;k<=T;k++) { memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); memset(c,0,sizeof(c)); memset(d,0,sizeof(d)); scanf("%s%s",a,b); len1=strlen(a); len2=strlen(b); len=len1>len2?len1:len2; for(i=len1-1,j=0;i>=0;i--,j++) { c[j]=a[i]-'0'; } for(i=len2-1,j=0;i>=0;i--,j++) { d[j]=b[i]-'0'; } int temp=0; for(i=0;i<=len;i++) { int sum=d[i]+c[i]+temp; d[i]=sum%10; temp=sum/10; } for(i=len;i>=0;i--) { if(d[i]!=0) break; } printf("Case %d:\n%s + %s = ",k,a,b); for(j=i;j>=0;j--) printf("%d",d[j]); if(k==T) printf("\n"); else printf("\n\n"); } return 0; }
相关文章推荐
- cache 一致性
- 使用类的静态字段和构造函数,我们可以跟踪某个类所创建对象的个数。请写一个类,在任何时候都可以向它查询“你已经创建了多少个对象?”。
- Java学习之Hessian通信基础
- Reverse Nodes in k-Group
- (原创)c#学习笔记03--变量和表达式03--变量02--变量的命名
- nginx预防蜘蛛爬虫处理
- 初识xml
- Java多线程相关知识
- 图像融合算法(感应篇)
- html多个水平并列组件自适应等高的做法
- Mac终端出现“terminal pointer being freed was not allocated…”解决办法
- C语言基础(二)常量、变量、类型转换、转义符、printf、scanf
- C++ operator关键字(重载操作符)
- C语言基础(一)编译、数据类型、注释、代码块
- 《大道至简》第三章 读后感
- java编程思想-复用类(2)
- 测试对象管理—共享对象库
- 基于BlueZ的C语言蓝牙编程
- Java字段初始化的规律
- Mvc Html.BeginForm 方式提交Form前验证