hdu--1002 A + B Problem II

A + B Problem II

:Time Limit 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 313981    Accepted Submission(s): 60834

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 utInp

21 2112233445566778899 998877665544332211

 

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110
代码：
#include<stdio.h>#include<stdlib.h>#include<string.h>int max(int x,int y){int z=x>y?x:y;return z;}int main(){int lena,lenb,n,t=0,numa[1100],numb[1100];char a[1100],b[1100];scanf("%d",&n);while(n--){getchar();scanf("%s%s",a,b);memset(numa,0,sizeof(numa));memset(numb,0,sizeof(numb));lena=strlen(a);lenb=strlen(b);for(int i=0;i<lena;i++){numa[lena-1-i]=a[i]-'0';}for(int i=0;i<lenb;i++){numb[lenb-1-i]=b[i]-'0';}int z=max(lena,lenb);for(int i=0;i<z;i++){numa[i]=numa[i]+numb[i];if(numa[i]>9){numa[i]=numa[i]-10;numa[i+1]++;}}t++;printf("Case %d:\n",t);printf("%s + %s = ",a,b);if(numa[z]==0){for(int i=z-1;i>=0;i--){printf("%d",numa[i]);}printf("\n");}else{for(int i=z;i>=0;i--){printf("%d",numa[i]);}printf("\n");}if(n!=0)  printf("\n");}return 0;}