大数专题A - A + B Problem II_a - a + b problem ii c-优快云博客

本文链接：https://blog.youkuaiyun.com/Cworld2017/article/details/75433688

本文提供了一种处理大整数相加的C语言程序实现方案。该方案绕过了使用标准整数类型的限制，允许处理长度可达1000位以上的数字。通过字符串输入大整数，并转换为字符数组进行逐位运算，最终输出加法结果。

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#include <stdio.h>
#include <string.h>
main()
{
	char a[1010],b[1010];int T,k,aa[1010],bb[1010],cc[1010],lena,lenb,lenc,i,x,flag=0;
	////freopen("1.txt","r",stdin);
	//freopen("2.txt","w",stdout);
	scanf("%d",&T);
	for(k=1;k<=T;k++)
	{
		memset(a,'\0',sizeof(a));memset(b,'\0',sizeof(b));
		scanf("%s%s",&a,&b);lena=strlen(a);lenb=strlen(b);
		memset(aa,0,sizeof(aa));memset(bb,0,sizeof(bb));memset(cc,0,sizeof(cc));
		for(i=0;i<lena;i++)aa[lena-i-1]=a[i]-48;//把lena-i-1改成lena-i就会报错
		for(i=0;i<lenb;i++)bb[lenb-i-1]=b[i]-48;
		lenc=0;x=0;
		while(lenc<lena||lenc<lenb)
		{
			cc[lenc]=aa[lenc]+bb[lenc]+x;
			x=cc[lenc]/10;
			cc[lenc]%=10;
			lenc++;
		}
		cc[lenc]=x;
		if(cc[lenc]==0)lenc--;
		if(flag)printf("\nCase %d:\n",k);
		if(!flag){flag=1;printf("Case %d:\n",k);}
		printf("%s + %s = ",a,b);
		for(i=lenc;i>=0;i--)printf("%d",cc[i]);
		putchar('\n');
	}
	return 0;
}

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