大数加法

A + B Problem II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 408777    Accepted Submission(s): 79219 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 21 2112233445566778899 998877665544332211   Sample Output Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110   Author Ignatius.L   Recommend We have carefully selected several similar problems for you:  1004 1003 1008 1005 1020    Statistic | Submit | Discuss | Note

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/*
	大数加法采用的是模拟的思想，就是利用数组来储存一个数的每一位 
*/
#include
#include
#include
using namespace std;
const int MAXN=10000;
char s1[MAXN],s2[MAXN];
int n1[MAXN],n2[MAXN],sum[MAXN];
int main()
{
	int T;
	scanf("%d",&T);
	for(int k=1;k<=T;k++)
	{
		scanf("%s %s",s1,s2);//用字符串来储存两个大数 
		memset(n1,0,sizeof(n1));//初始化数组，让它们全为0 
		memset(n2,0,sizeof(n2));
		memset(sum,0,sizeof(sum));//初始化保存结果的数组 
		int len1=strlen(s1);//第一个数的长度 
		int len2=strlen(s2);//第二个数的长度 
		int j=0;
		for(int i=len1-1;i>=0;i--)//将第一个数的每一位都逆序赋值给第一个数组 
			n1[j++]=s1[i]-'0';
		j=0;	
		for(int i=len2-1;i>=0;i--)
			n2[j++]=s2[i]-'0';
		int len=len1>len2?len1:len2;//找出来两个数中比较长的那个数 
		int pre=0;//用来保存进位 
		for(int i=0;i9)//保存最高位的是sum[len-1] ，如果大于9 ，结果的位数要加一 
		{
			sum[len]=pre/10;//取高位 
			len++;//位数 +1
		}
		int t=len;
		for(int i=len-1;i>=0;i--)	//去掉前置0 
		{
			if(sum[i]==0)	t--;//像 0001 + 3 这种，结果应该输出 4 而不是0004 
			else			break;
		}
		printf("Case %d:\n%s + %s = ",k,s1,s2);
		for(int i=t-1;i>=0;i--)		
			printf("%d",sum[i]%10);
		printf("\n");
		if(k!=T)	printf("\n");//最后一组数据不要空行 
	 } 
	return 0;
}