NYOJ 103 A+B Problem II

青旅半醒 2022-08-25 01:43 24阅读 0赞

## 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

思路：可以用字符串，也可以用整型数组。

#include <stdio.h>
    #include <string.h>
    
    void reverse(char *s)
    {
    	char *last,temp;
    	for (last = s; *last != '\0'; last++);
    	last--;
    	while (s<last)
    	{
    		temp = *s;
    		*s++ = *last;
    		*last-- = temp;
    	}
    }
    
    int main()
    {
    	int m,t = 1;
    	char a[1010],b[1010],c[1010];
    	scanf("%d",&m);
    	while (m--)
    	{
    		int i,j;
    		int carry = 0;
    		scanf("%s %s",a,b);
    		reverse(a);
    		reverse(b);
    		printf("Case %d:\n",t);
    		t++;
    		i = strlen(a) < strlen(b)?strlen(a):strlen(b);
    		for (j=0; j<i; j++)
    		{
    			c[j] = a[j] + b[j] - '0' - '0' + carry;
    			carry = c[j] / 10;
    			c[j] = c[j] % 10 + '0';
    		}
    
    		while (j<strlen(a))
    		{
    			c[j] = a[j] - '0' + carry;
    			carry = c[j] / 10;
    			c[j] = c[j] % 10 + '0';
    			j++;
    		}
    		while (j<strlen(b))
    		{
    			c[j] = b[j] - '0' + carry;
    			carry = c[j] / 10;
    			c[j] = c[j] % 10 + '0';
    			j++;
    		}
    		if (carry != 0)
    		{
    			 c[j] = carry + '0';
    			 j++;
    		}
    		c[j] = '\0';
    		reverse(a);
    		reverse(b);
    		reverse(c);
    		printf("%s + %s = %s\n",a,b,c);
    	}
    	return 0;
    }

#include <stdio.h>
    #include <string.h>
    
    
    #define N 1001
    
    
    void charstoint(int a[], char s[])
    {
    	int i;
    	int len = strlen(s);
    	memset(a,0,4*N);   //整型占4个字节
    
    
    	for (i=0; i<len; i++)
    	{
    		//不考虑负数与非数字字符的情况
    		a[len-1-i] = s[i] - '0'; //将每一个字符转化为数字逆序存放在整型数组中，例如字符串 123456 -> a[0] = 6,a[1] = 5, ...a[5] = 1;
    	}
    }
    
    
    void add(int a[], int b[], int c[])
    {
    	int i;
    
    
    	memset(c,0,4*(N+1)); 
    
    
    	for (i=0; i<N; i++)
    	{
    		c[i] = a[i] + b[i]; 
    	}
    
    
    	for (i=0; i<N+1; i++)  //处理进位
    	{
    		c[i+1] += c[i] / 10;
    		c[i] = c[i] % 10;
    	}
    }
    
    
    int main()
    {
    	int m,count = 0;
    	scanf("%d",&m);
    	while (m--)
    	{
    	int i,j = N;
    
    
    	int a[N],b[N],c[N+1];
    
    
    	char s1[N],s2[N];
    	scanf("%s %s",s1,s2);
    	count++;
    	printf("Case %d:\n",count);
    	printf("%s + %s = ",s1,s2);
    	charstoint(a,s1);
    	charstoint(b,s2);
    
    
    	add(a,b,c);
    
    
    	while(c[j]==0)   //去掉多余的0
    	{
    		j--;
    	}
    
    
    	for (i=j; i>=0; --i)  //打印结果
    	{
    		printf("%d",c[i]);
    	}
    
    
    	printf("\n");
    	}
    
    
    	return 0;
    }