HDOJ 1002 A + B Problem II 大整数相加高效率版的C语言实现

A + B Problem II

[b]Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 255206 Accepted Submission(s): 49245[/b]

Problem Description
I have a verysimple problem for you. Given two integers A and B, your job is to calculatethe Sum of A + B.

Input
The firstline of the input contains an integer T(1<=T<=20) which means the numberof test cases. Then T lines follow, each line consists of two positiveintegers, A and B. Notice that the integers are very large, that means youshould not process
them by using 32-bit integer. You may assume the length ofeach integer will not exceed 1000.

Output
For each testcase, 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 intthe 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

#include<stdio.h>
#include<string.h>

int main()
{
int len_max,len_min,len_str_a,len_str_b,t,i,j,c,temp;
char str_a[1001],str_b[1001],str_c[1001];
char *len_max_str;
scanf("%d",&t);
for(j=1; j<=t; j++)
{
//输入两个加数的字符串并记录长度
scanf("%s",str_a);
len_str_a = strlen(str_a);
scanf("%s",str_b);
len_str_b = strlen(str_b);
//记录最大长度、最小长度、最大长度字符串首地址
if(len_str_a>len_str_b)
{
len_max = len_str_a;
len_max_str = str_a;
len_min = len_str_b;
}
else
{
len_max = len_str_b;
len_max_str = str_b;
len_min = len_str_a;
}

c=0; //进位初始为0
for(i=0; i<len_max; i++) //从个位开始相加
{
//得到当前位相加的结果
if(i<len_min) //如果两个数都存在该位，那么当前位的结果就是就把两数的该位以及进位数相加
{
temp=(str_a[len_str_a-1-i]-'0')+(str_b[len_str_b-1-i]-'0')+c;
}
else  //如果只有长度较长的数存在该位，那么当前位的结果就是长度较长的数该位的值加上进位数
{
temp = len_max_str[len_max-1-i]-'0'+c;
}
//记录当前位最终结果
str_c[len_max-1-i] = temp%10+'0';
//记录进位数
c=temp/10;
}
str_c[len_max] = '\0';//给结果的字符串添加结束标志
//输出结果
printf("Case %d:\n%s + %s = ",j,str_a,str_b);
if(c!=0) //如果最高位还存在进位，输出的时候就多个1（两数相加进位最多为1）
{
printf("1%s",str_c);
}
else
{
printf("%s",str_c);
}
printf("\n");
if(j<t) //题目要求在每两个输出之间价格空行，并不是每个输出都加（最后一行不加）
{
printf("\n");
}
}
return 0;
}