hdoj1002--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 should not 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 the equation. Output a blank line between two test cases.

[align=left]Sample Input[/align]

2
1 2
112233445566778899 998877665544332211

[align=left]Sample Output[/align]

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

简单题：大数的运算
注意格式（Case的首字母大写、各种空格、每一行之间有空行，最后一行没有空行）！

给出几组测试数据:

6
1 2
1 0
9999 1
1 999999
5555 4445
112233445566778899 998877665544332211

java code:

import java.util.*;
import java.io.*;
import java.math.*;

class Main
{
public static BigInteger plus(BigInteger a, BigInteger b) {
BigInteger c;
c = a.add(b);
return c;
}

public static void main(String args[])
{
Scanner cin = new Scanner(System.in);
int T, i;
BigInteger a,b;
T = cin.nextInt();
i = 1;
while((T--) > 0) {
a = cin.nextBigInteger();
b = cin.nextBigInteger();
System.out.println("Case "+ i +":");
System.out.println(a + " + "+ b + " = "+ plus(a, b));
i++;
if(T > 0)
System.out.println();
}
}
}

C code

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
void solve()
{
int n,i,j,k,flag,t,cas,L;
char a[1001],b[1001],c[1002];
scanf("%d",&n);
getchar();
cas=1;
while(n--)
{
flag=0;
memset(a,'\0',sizeof(a));
memset(b,'\0',sizeof(b));
memset(c,'\0',sizeof(c));
scanf("%s",a);
scanf("%s",b);
printf("Case %d:\n",cas);
printf("%s",a);
printf(" + ");
printf("%s",b);
printf(" = ");
strrev(a);
strrev(b);
k=i=0;
L=(strlen(a)>strlen(b)?strlen(b):strlen(a));
while(i<L)
{
t=(a[i]-'0')+(b[i]-'0')+flag;
flag=(t>=10?1:0);
c[k++]=t%10+'0';
i++;
}
if(a[i]=='\0')
{
while(b[i]!='\0') {
t=b[i++]-'0'+flag;
c[k++]=t%10+'0';
flag=(t>=10?1:0);
}
}
else
{
while(a[i]!='\0') {
t=a[i++]-'0'+flag;
c[k++]=t%10+'0';
flag=(t>=10?1:0);
}
}
if(flag) c[k]='1';
else k--;
while(k>=0)
{
printf("%c",c[k]);
k--;
}
printf("\n");
if(n>0) printf("\n");
cas++;
}
}

int main()
{
solve();
return 0;
}