1002 1002 A + B Problem II大数相加

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A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 317528 Accepted Submission(s): 61692

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

21 2112233445566778899 998877665544332211

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

Author

Ignatius.L

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1002

大数相加思路非常简单，只要用数组将大数的每一位都存起来，注意进位。。。。。

AC代码：

#include<stdio.h>#include<string.h>#define Max 101void print(char sum[]);void bigNumAdd(char a[],char b[],char sum[]);int main(){char a[Max];char b[Max];char sum[Max];gets(a);gets(b);bigNumAdd(a,b,sum);print(sum);return 0;}void bigNumAdd(char a[],char b[],char sum[]){int i=0;int c=0;//表示进位          //初始化,对以后位运算有很大帮助！char m[Max]={0};char n[Max]={0};memset(sum,0,Max*sizeof(char)); //这里不能写成memset(sum,0,sizeof(sum));原因见注意事项1//字符串反转且字符串变数字int lenA=strlen(a);int lenB=strlen(b);for (i=0;i<lenA;i++){m[i]=a[lenA-i-1]-'0';}for (i=0;i<lenB;i++){n[i]=b[lenB-i-1]-'0';}//位运算for (i=0;i<lenA||i<lenB;i++){sum[i]=(m[i]+n[i]+c)%10+'0';//得到末位c=(m[i]+n[i]+c)/10;//得到进位}}void print(char sum[]){int i=0;int j=0;int len = strlen(sum);for (i=len-1;sum[i]==0;i--); //找到第一个不为零的位置，方便输出for (j=i;j>=0;j--){printf("%c",sum[j]);}}

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