大数A+B问题

原理

        大数运算的原理其实就是模拟人工计算(注记:再考虑是否有其他算法。注记日期:2017.3.19),人工加法计算步骤如下:

    1.将两个操作数(operand)位数对齐。

    2.从最低位开始,计算两个操作数每位的总和再加上进位数(第一位数为0)的结果,保留其个位数。

    3.若上述的结果大于10,进位数置为1,否则为0。(每位数相加的结果不会超过 9 + 9 = 18 , 因此进位数只有1和0)

    4.重复上述步骤2、3,直到计算完两个操作数其中那个位数小的最高位。

我们来看一道例题:

南阳oj-A+B Problem II


A+B Problem II

时间限制: 3000 ms  |  内存限制: 65535 KB
难度: 3
描述

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

A,B must be positive.

输入
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.
输出
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.
样例输入
2
1 2
112233445566778899 998877665544332211
样例输出
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

注意:此题有一个细节问题就是高位不能为0,举个例子012+011不能等于033,而结果为33
现在我们来看一下代码的实现:

#include
#include
#include
#include
#define MAX 1010
using namespace std;
char str1[MAX],str2[MAX];
int a[MAX],b[MAX],c[MAX]; 
int main()
{
	int t;
	int kase=0;
	cin>>t;
	while(t--)
	{
		memset(a,0,sizeof(a));//对数组初始化 
		memset(b,0,sizeof(b));
		memset(c,0,sizeof(c));
		cin>>str1>>str2; 
		printf("Case %d:\n",++kase);
		printf("%s + %s = ",str1,str2);
		int len1=strlen(str1);
		int len2=strlen(str2);
		for(int i=len1-1,j=0;i>=0;i--)
		{
			a[j++]=str1[i]-'0';
		}
		for(int i=len2-1,j=0;i>=0;i--)
		{
			b[j++]=str2[i]-'0';
		}
		for(int i=0;i=10)
			{
				c[i]=c[i]%10;//满十进一 
				c[i+1]++;
			}
		}
		int j; 
		for(j=MAX-1;c[j]==0;j--);//避免高位等于0 
		if(j<0)
		cout<<0;
		else
		{
			for(;j>=0;j--)
			cout<




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