杭电OJ题 1002 A + B Problem II 解题报告

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本文介绍了一个简单的A+B问题，即给定两个整数A和B，计算它们的和。该问题是通过处理大数加法来实现的，涉及到字符串处理和逐位相加的算法。文章提供了完整的C语言代码实现，包括输入输出格式、进位处理等细节。

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A + B Problem II

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

Total Submission(s): 141652 Accepted Submission(s): 26876

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

——————————————————————————————————————

简单的大数加法问题，注意进位

/****************************
 *Name:A + B Problem II.c
 *Tags:ACM BigNumber
 ****************************/
#include <stdio.h>
#include <string.h>

int main()
{
      char num1[1002], num2[1002], res[1003];
      int T, c, l1, l2, i, j, count = 0;
      scanf("%d", &T);
      while(T--) {
	    scanf("%s %s", num1, num2);
	    l1 = strlen(num1);
	    l2 = strlen(num2);
	    for(i = 0; i < 1003; i++) {
		  res[i] = '0';
	    }
	    for(i = l1-1, j = 0; i >= 0; i--) {
		  res[j++] = num1[i];
	    }
	    for(i = l2-1, j = 0; i >= 0; i--) {
		  res[j++] += (num2[i]-'0');
	    }
	    j = l1>l2?l1:l2;
	    c = 0;
	    for(i = 0; i < j; i++) {
		  res[i] += c;
		  c = (res[i]-'0') / 10;
		  res[i] = (res[i]-'0') % 10+'0';
	    }
	    count++;
	    printf("Case %d:\n%s + %s = ",count ,num1, num2);
	    if(c) {
		  printf("%c", c+'0');
	    }
	    for(i = j-1; i >= 0; i--) {
		  printf("%c", res[i]);
	    }
	    printf("\n");
	    if(T) {
		  printf("\n");
	    }
      }
      return 0;
}