ACM1002：A + B Problem II

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
--------------------------------------------------------------------------------------------------------------------------

 

//高精度加法，注意格式

#define _CRT_SECURE_NO_WARNINGS

#include<stdio.h>

#include<stdlib.h>

#include<string.h>

#include<memory.h>

#define SIZE 1005

int main()

{

	int i, n;

	int j, k, m;

	int index;

	int lengthA, lengthB;

	char a[SIZE];

	char b[SIZE];

	int ans[SIZE];

	int carryBit;    //进位

	//freopen("F:\\input.txt","r",stdin);

	scanf("%d", &n);

	for(i = 0; i < n; i++)

	{

		scanf("%s%s", a, b);

		memset(ans, 0, sizeof(ans));

		index = 0;

		carryBit = 0;

		lengthA = strlen(a);

		lengthB = strlen(b);

		lengthA--;

		lengthB--;

		while ((lengthA >= 0) && (lengthB >= 0))

		{

			//将char转换为int,并计算和，如果ans大于10，则carryBit为1，否则为0

			ans[index] = (a[lengthA] - '0') + (b[lengthB] - '0') + carryBit;

			if (ans[index] >= 10)

			{

				carryBit = ans[index] / 10;

				ans[index] %= 10;

			}

			else

				carryBit = 0;

			index++;

			lengthA--;

			lengthB--;

		}

		//如果A的长度大于B，则将A与carryBit相加

		while (lengthA >= 0)

		{

			ans[index] = (a[lengthA] - '0') + carryBit;

			if (ans[index] >= 10)

			{

				carryBit = ans[index] / 10;

				ans[index] %= 10;

			}

			else

				carryBit = 0;

			index++;

			lengthA--;

		}

		//如果B的长度大于A，则将B与carryBit相加

		while (lengthB >= 0)

		{

			ans[index] = (b[lengthB] - '0') + carryBit;

			if (ans[index] >= 10)

			{

				carryBit = ans[index] / 10;

				ans[index] %= 10;

			}

			else

				carryBit = 0;

			index++;

			lengthB--;

		}

		if (carryBit != 0)

		{

			ans[index]++;

			index++;

		}

		//去掉多余的零

		while (ans[index] == 0)

			index--;

		printf("Case %d:\n", i + 1);

		printf("%s + %s = ", a, b);

		//两个零字符串相加输出为0

		if (index == -1)

			printf("0");

		else

		{

			for (m = index; m >= 0; m--)

				printf("%d", ans[m]);

		}

		printf("\n");

		if (i != n - 1)

			printf("\n");

	}

	//freopen("con", "r", stdin);

	//system("pause");

	return 0;

}