计算两个大整数的和

我会带着你远行 2022-08-03 17:29 202阅读 0赞

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
    
    
    一开始我以为用long long型的数字就可以满足条件了，但是我编译器不通过，我就试了哈提交上去，但是答案显示wa .
    虽然对于很大的数据不对，但是对于long型以内的数据还是满足的，所以还是先把代码贴上来，后面贴满足真正意义上的大数据的代码
    
    
       
        #include <stdio.h>
    
    
    long len(long num1,long num2)
    {
    	long sum = 0 , sum_tmp = 1 , sum_two = 0;
    	while(num1 != 0 && num2 != 0)
    		{
    			sum_two = (num1%10) + (num2%10);
    			sum += sum_tmp*sum_two;
    			sum_tmp *= 10;
    			num1 /= 10;
    			num2 /= 10;
    		}
    	if(num1 != 0)
    	{
    		while(num1 != 0)
    		{
    			sum_two = num1%10;
    			sum += sum_tmp*sum_two;
    			sum_tmp *= 10;
    			num1 /= 10;
    		}
    		
    	}
    	else if(num2 != 0)
    	{
    		while(num2 != 0)
    		{
    			sum_two = num2%10;
    			sum += sum_tmp*sum_two;
    			sum_tmp *= 10;
    			num2 /= 10;
    		}
    		
    	}
    	return sum;
    }
    
    int main()
    {
    	long n , a , b ,sum = 0;
    	int i ;
    	scanf("%d",&n);
    	for(i = 0 ; i < n ; i ++)
    	{
    		scanf("%lld%lld",&a,&b);
    		
    		sum = len(a,b);
    
    		printf("case %d:\n",i+1);
    		if(i != n-1)
    		{	
    			printf("%lld+%lld=%lld\n\n",a,b,sum);
    		}
    		else if(i == n-1)
    			printf("%lld+%lld=%lld\n",a,b,sum);
    	}
    	return 0 ;
    } 
       
        
    
       
        下面贴上真正意义上符合大数据加法的代码： 
       
        
    
       
        #include<stdio.h>
    #include<string.h>
    
    void reverse( char *s )        /*将字符串逆置*/
    {
        int length;
        int i = 0;
        char temp;
        length = strlen( s );
        while( i < length - i - 1 )
        {
            temp = s[i];
            s[i] = s[length - i - 1];
            s[length - i - 1] = temp;
            i++;
        }
    }
    
    void AddBigNum( char* s1, char* s2, char* result )
    {
        int len1 = strlen( s1 );
        int len2 = strlen( s2 );
        int acc = 0, temp, i;        /*acc为进位标记*/
        if( s1 == NULL || s2 == NULL || result == NULL )
        {
            return;
        }
        reverse( s1 );
        reverse( s2 );
        for( i = 0; i < len1 && i < len2; i++ )
        {
            temp = s1[i] - '0' + s2[i] - '0' + acc;        /*计算每位的实际和*/
            result[i] = temp % 10 + '0';        /*通过求余数来确定每位的最终值*/
            if( temp >= 10 )        /*通过这个if..else..条件来判断是否有进位，并设置进位值*/
                acc = 1;
            else
                acc = 0;
        }
        if( i < len1 )        /*两个加数位数不同*/
        {
            for( ; i < len1; i++ )
            {
                temp = s1[i] - '0' + acc;        /*依旧要考虑进位，比如9999 + 1的情况*/
                result[i] = temp % 10 + '0';
                if( temp >= 10 )        
                    acc = 1;
                else
                    acc = 0;
            }
        }
        if( i < len2 )
        {
            for( ; i < len2; i++ )
            {
                temp = s2[i] - '0' + acc;
                result[i] = temp % 10 + '0';
                if( temp >= 10 )
                    acc = 1;
                else
                    acc = 0;
            }
        }
    
        if( acc == 1 )        /*考虑如:123 + 911 = 1034的情况，如果不增加这个条件会得到结果为034，进位被舍弃*/
    
            result[i++] = '1';
    
        result[i] = '\0';
        reverse( result );
    }
    
    void main()
    {
        char s1[1000] = {0};
        char s2[1000] = {0};
    	char result[1002];
    	int n,i;
    	scanf("%d",&n);
    	for(i = 0 ; i < n ; i ++)
    	{
    		scanf("%s%s",s1,s2);
    		printf("case %d:\n",i+1);
    		printf("%s + %s = ",s1,s2);
    		AddBigNum( s1, s2, result );
    		if(i != n-1)
    		{
    			printf( "%s\n\n", result );
    		}
    		else if(i == n-1)
    			printf( "%s\n", result );
    	}
    
    } 结果还是wa...好难过