A + B Problem II（高精度）

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 188211    Accepted Submission(s): 35956

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

 

 

Author

 

Ignatius.L

    题意：

    给出 T （1 ~ 20）个case，后给出两个数a,b，每个数的长度最多1000位，求两个数加和按要求格式输出。

    

    思路：

    输入的顺序是从高位向地位输入，故计算的时候从 n -> 0 计算，并且考虑进位情况。考虑到数的输入输出情况，故开 char 型数组比较方便，计算的时候相应减去字符 0 即可。

    两个数的长度可能不一致，故先对其，前补0再计算，输出的时候不输出前导0，每个case之间要空一行，最后一个case不需要空行，大概注意这些就可以AC了。

 

    AC：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

int main()
{
    int n,time = 0;
    scanf("%d",&n);
    while(n--)
    {
        int CIN = 0,len_a,len_b,len_sum;
        char a[1005],b[1005],sum[1005],temp[1005],temp_a[1005],temp_b[1005];
        time++;
        scanf("%s%s",a,b);
        len_a = strlen(a);
        len_b = strlen(b);
        len_sum = max(len_a,len_b);
        strcpy(temp_a,a);
        strcpy(temp_b,b);
        if(len_a > len_b)
        {
            int idex_b = 0,i;
            for(i = 0;i < len_a - len_b;i++)
                temp[i] = '0';
            for(i;i < len_a;i++,idex_b++)
                temp[i] = b[idex_b];
            for(i = 0;i < len_a;i++)
                b[i] = temp[i];
            b[i] = '\0';
        }

        if(len_b > len_a)
        {
            int idex_a = 0,i;
            for(i = 0;i < len_b - len_a;i++)
                temp[i] = '0';
            for(i;i < len_b;i++,idex_a++)
                temp[i] = a[idex_a];
            for(int i = 0;i < len_b;i++)
                a[i] = temp[i];
            a[i] = '\0';
        }
        for(int i = len_sum - 1;i >= 0;i--)
        {
            int ans = (a[i] + b[i] - '0' - '0' + CIN);
            sum[i] = ans % 10 + '0';
            CIN = ans / 10;
        }
        sum[len_sum] = '\0';
        printf("Case %d:\n",time);
        printf("%s + %s = ",temp_a,temp_b);
        if(CIN) printf("%d",CIN);
        puts(sum);
        if(n) printf("\n");
    }
    return 0;
}