P1001: A + B Problem II

╰+哭是因爲堅強的太久メ 2022-12-04 01:21 124阅读 0赞

## P1001: A + B Problem II ##

题目描述  
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.  
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. Output a blank line between two test cases.  
样例输入

2
    1 2
    112233445566778899 998877665544332211

样例输出

Case 1:
    1 + 2 = 3
    
    Case 2:
    112233445566778899 + 998877665544332211 = 1111111111111111110

#include<iostream>
    #include<cstring>//或者#include<string.h>,C++这两个都行，而C只能是.h
    using namespace std;
    
    int JiaFa(string num1, string num2, char num[])
    { 
    	int len1 = num1.length(), len2 = num2.length();
    	int i = len1-1,j=len2-1,k=0;
    	int yuxiang = 0;
    	while(i>=0||j>=0)
    	{ 
    		if (i >=0 && j >= 0)
    		{ 
    			num[k++] = ((num1[i] - '0') + (num2[j] - '0') + yuxiang) % 10 + '0';
    			yuxiang = ((num1[i] - '0') + (num2[j] - '0') + yuxiang) / 10;
    			i--;
    			j--;
    		}
    		else if (i <0)
    		{ 
    			num[k++] = ((num2[j] - '0') + yuxiang) % 10+'0';
    			yuxiang= ((num2[j] - '0') + yuxiang) / 10;
    			j--;
    		}
    		else
    		{ 
    			num[k++] = ((num1[i] - '0') + yuxiang) % 10 + '0';
    			yuxiang = ((num1[i] - '0') + yuxiang) / 10;
    			i--;
    		}
    	}
    	if (yuxiang)
    	{ 
    		num[k] = yuxiang + '0';
    		num[k + 1] = '\0';
    	}
    	else
    		num[k] = '\0';
    	{ 
    		int i = 0, j = strlen(num)-1;
    		for (; i < j; i++, j--)
    		{ 
    			char temp = num[i];
    			num[i] = num[j];
    			num[j] = temp;
    		}
    	}
    	return 1;
    }
    
    
    int main()
    { 
    	string num1[100], num2[100];
    	char num[100][10000];
    	int n,i;
    
    	cin >> n;
    	for (i = 0; i < n; i++)
    		cin >> num1[i] >> num2[i];
    	for (i = 0; i < n; i++)
    	{ 
    		JiaFa(num1[i],num2[i],num[i]);
    		cout << "Case " << i+1 << ":" << endl;
    		cout << num1[i] << " + " << num2[i] << " = " << num[i] << endl;
    		cout << endl;
    	}
    	return 0;
    }