HDU 1002 A A + B Problem II （大数问题）

原题代号：HDU 1002

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1002

原题描述：

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

 

题目大意：完成长度在1000以内的任意两数相加 （转换成字符串处理）

解法：

 

# include <stdio.h>
# include <string.h>
# include <stdlib.h>
# include <iostream>
# include <fstream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <math.h>
# include <algorithm>
using namespace std;
# define pi acos(-1.0)
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define For(i,n,a) for(int i=n; i>=a; --i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define Fo(i,n,a) for(int i=n; i>a ;--i)
typedef long long LL;
typedef unsigned long long ULL;

char a[1000+5],b[1000+5],c[1000+5];

int main()
{
    int t,sum=0;
    cin>>t;
    while(t--)
    {
        if(sum)cout<<endl;
        char str1[1000+5],str2[1000+5];
        mem(a,'\0');
        mem(b,'\0');
        mem(c,'\0');//因为多组数据，所以清零
        cin>>str1;
        for(int i=sizeof(a),j=strlen(str1)-1; j>=0; j--,i--)
            a[i]=str1[j]-48;
        cin>>str2;
        for(int i=sizeof(b),j=strlen(str2)-1; j>=0; j--,i--)//将字符串放在数组末尾，并且将字符转换成ASCLL码所对应的数字
            b[i]=str2[j]-48;
        int num=0,i;
        for(i=sizeof(a); i>0; i--)
        {
            c[i]=a[i]+b[i]+num;
            num=c[i]/10;
            c[i]%=10;
        }
        printf("Case %d:\n%s + %s = ",++sum,str1,str2);
        for(i=0;i<=10005;i++)if(c[i])break;
        for(int j=i; j<=sizeof(c); j++)
            printf("%d",c[j]);//将大数表示出来
        cout<<endl;
    }
    return 0;
}