A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 392274 Accepted Submission(s): 75946

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

这个方法没有ac 
import java.util.Scanner;
public class P1002 {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
while(sc.hasNext()){
int t=sc.nextInt();
long[][] a=new long[t][2];
long[] sum=new long[t];
for(int i=0;i<t;i++){
for(int j=0;j<2;j++){
a[i][j]=sc.nextLong();
}
}
for(int i=0;i<t;i++){
sum[i]+=a[i][0]+a[i][1];
}
for(int i=0;i<t;i++){
System.out.println("Case"+(i+1)+":");
System.out.println(a[i][0]+"+"+a[i][1]+"="+sum[i]);
System.out.println();
}
}
}
}

大数初步运用

/*直接运用java中的BigDecimal方法,ac/
import java.math.BigDecimal;
import java.util.Scanner;
public class P1002 {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
while(sc.hasNext()){
int n=sc.nextInt();
int num=0;
for(int i=0;i<n;i++){
BigDecimal a=sc.nextBigDecimal();
BigDecimal b=sc.nextBigDecimal();
BigDecimal c=a.add(b);
num++;
System.out.println("Case "+num+":");
System.out.println(a.toString()+" + "+b.toString()+" = "+c.toString());
if(i<n-1){
System.out.println();
}
}
}
}
}