hdu1002 A + B Problem II(大数题)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 230247 Accepted Submission(s): 44185


[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

题目大意：题意很容易理解，具体就不解释了，主要就是要解决大数的问题。

题目思路：如果会java的话，可以轻松AC。其他的小伙伴们只能用最笨的方法解决。我们用一个数字将数字倒过来存下，无论是乘法还是加法，这是最好的解决办法。

下面附上两个代码。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int main ()
{
char a[8000],b[8000];
int na[8000],nb[8000],sum[8000],pre,flag=1;
int t;
scanf("%d",&t);
while (t--)
{
memset(sum,0,sizeof(sum));
memset(na,0,sizeof(na));
memset(nb,0,sizeof(nb));
scanf("%s%s",a,b);
pre=0;
int lena=strlen(a);
int lenb=strlen(b);
for (int i=0; i<lena; i++)
na[lena-1-i]=a[i]-'0';
for (int j=0; j<lenb; j++)
nb[lenb-1-j]=b[j]-'0';
int lenx=lena>lenb?lena:lenb;
for (int k=0; k<lenx; k++)
{
sum[k]=na[k]+nb[k]+pre/10;
pre=sum[k];
}
while (pre>9)
{
sum[lenx]=pre/10%10;
lenx++;
pre/=10;
}
printf ("Case %d:\n",flag++);
printf ("%s + %s = ",a,b);
for (int i=lenx-1; i>=0; i--)
{
printf ("%d",sum[i]%10);
}
printf ("\n");
if (t)
printf ("\n");
}

return 0;
}

java代码。

import java.util.*;
import java.math.*;
public class Main {
public static void main(String[] args) {
Scanner sc=new Scanner (System.in);
int l=sc.nextInt();
for(int i=1;i<=l;i++){
if(i!=1) System.out.println();
BigInteger a,b;
a=sc.nextBigInteger();
b=sc.nextBigInteger();
System.out.println("Case "+i+":");
System.out.println(a+" + "+b+" = "+a.add(b));
}
}
}