HDU 1002 大数加法

A + B Problem II

[align=left]Problem Description[/align]I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B. [align=left]Input[/align]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 shouldnot process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000. [align=left]Output[/align]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 theequation. Output a blank line between two test cases. [align=left]Sample Input[/align]

21 2112233445566778899 998877665544332211 [align=left]Sample Output[/align]

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

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;char a[1050],b[1050],c[1050],d[1050],e[1050];int main(){int t,j,i,lena,lenb,lenc,num;scanf("%d",&t);for(i=1;i<=t;i++){scanf("%s%s",a,b);memset(c,0,sizeof(c));memset(d,0,sizeof(d));memset(e,0,sizeof(e));lena=strlen(a);lenb=strlen(b);lenc=max(lena,lenb);num=0;for(j=0;j<=lena-1;j++){c[lena-j-1]=a[j]-'0';}for(j=0;j<=lenb-1;j++){d[lenb-j-1]=b[j]-'0';}for(j=0;j<=lenc;j++){e[j]=c[j]+d[j]+num;if(e[j]>9){e[j]-=10;num=1;}else{num=0;}}if(e[lenc]==0)lenc--;i==1?printf("Case %d:\n%s + %s = ",i,a,b):printf("\nCase %d:\n%s + %s = ",i,a,b);for(j=lenc;j>=0;j--){printf("%d",e[j]);}printf("\n");}}