HDU 1002 A + B Problem II(两个大数相加)

详细题目点击：http://acm.hdu.edu.cn/showproblem.php?pid=1002

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

21 2112233445566778899 998877665544332211

 

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

        分析：对于此题做法有两种：其一，使2字符串的中的字符数字减去'0'，逐个相加大于等于10的可以使本位减10，下一位自增1，后面的处理就非常简单了；其二，便是读入字符串后先让各个字符减'0'，一一对应存入整形数组中;之后再相加。对于2种方法大致是相同的，都要从后面向前加，逢十进位，以及数组初始化均要初始为0，一边方便运算。

#include <stdio.h>#include <stdlib.h>#include <string.h>char a[1001]; //开辟两个字符数组a、b，作为两个输入的大数char b[1001];char c[1002];int main(void){    int carry = 0, n, j;    int lena, lenb, i, lenc;    scanf("%d", &n);    for(j = 1; j <= n; j++)    {        memset(a, 0, 1001);        memset(b, 0, 1001);        memset(c, 0, 1002);        scanf("%s", a);        scanf("%s", b);        lena = strlen(a);        lenb = strlen(b);        for(lena--, lenb--, i = 0, carry = 0; (lena >= 0) && (lenb >= 0); lena--, lenb--, i++)        {            c[i] = a[lena]-'0' + b[lenb]-'0' + carry;            if((int)c[i] > 9)            {                c[i] = c[i] - 10 + '0';                carry = 1;            }            else            {                c[i] += '0';                carry = 0;            }        }        while(lena >= 0)        {            c[i] = c[i] + a[lena] + carry; //有可能加上carry后还可以向前进位            if(c[i] > '9')            {                c[i] -= 10;                carry = 1;            }            else                carry = 0;            i++;            lena--;        }        while(lenb >= 0)        {            c[i] = c[i] + b[lenb] + carry;            if(c[i] > '9')            {                c[i] -= 10;                carry = 1;            }            else                carry = 0;            i++;            lenb--;        }        lenc = strlen(c);        printf("Case %d:\n", j);        printf("%s + %s = ", a, b);        for(i = lenc-1; i >= 0; i--) //c数组中c[0]存放的是大数的最低位，c[lenc-1]存放的是大数的最高位            printf("%c", c[i]);        printf("\n");        if(j != n)            printf("\n");    }    return 0;}

参考资料：

1、(网上资料)hdu 1002 A + B Problem II 大整数相加