HDU 1002 A + B Problem II （大数加法）

题目链接

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

分析：

大数问题肯定是当作字符串来处理的，首先要把数的每个位数逆序存到一个int型的数组中，然后将这两个数组的对应位置上的数相加，如果有进位的话还应加上进位，最后输出的时候应该先找到首个不为0 的数，然后将后面的数输出来。

代码：

#include<iostream>
#include<math.h>
#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
    int n;
    scanf("%d",&n);
    getchar();
    for(int h=1; h<=n; h++)
    {
        char a[10000],b[10000];
        int c[10000],d[10000];
        int e[10000];
        int k,l;
        memset(c,0,sizeof(c));
        memset(d,0,sizeof(d));
      scanf(" %s %s",a,b);
        k=strlen(a);
        l=strlen(b);
        for(int i=0; i<k; i++)
            c[i]=a[k-i-1]-'0';
        for(int i=0; i<l; i++)
            d[i]=b[l-i-1]-'0';
        int op=0;
        for(int i=0; i<1000; i++)
        {
            e[i]=(c[i])+(d[i])+op;
            if(e[i]>=10)
            {
                e[i]=e[i]%10;
                op=1;
            }
            else op=0;
        }
        printf("Case %d:\n",h);
        printf("%s + %s = ",a,b);
        int i;
        for( i=999; i>=0; i--)
            if(e[i]!=0)break;
        for(int j=i; j>=0; j--)
            printf("%d",e[j]);
        printf("\n");
        if(h!=n)
            printf("\n");
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
    }
    return 0;
}