大数加法模拟-优快云博客

本文介绍了一种使用字符串模拟大数加法的方法，解决了输入数据超出32位整数限制的问题，并提供了完整的C语言实现代码。

D - A + B Problem II

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status

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.

正整数高精度1000位加法

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

思路：用字符串保存数据，模拟加法，从第位向高位循环操作。

失误：主要表现为思路不太清晰”，当遇到问题时无法解决又要从头考虑，浪费了时间

应大致确定思路，再从小处出发，模式化考虑，将问题分解为多个函数。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

int max(int la,int lb) //c的数学函数库中无此函数和min()
{
if(la>lb)
return la;
return lb;
}

int main()
{
int t,i,j,k=0,c[1111],num1[1111],num2[1111],la,lb,l,cnt;
char a[1111],b[1111];
scanf("%d",&t);

while(t--)
{
getchar(); //gets()函数可吸收一行字符包括\n 因此在gets之后如果再有字符型的输入函数则不用加getchar()

scanf("%s%s",a,b);// 但是scanf就要加了否则就会将回车当成下一个输入

la=strlen(a);
lb=strlen(b);
memset(num1,0,sizeof(num1));//数组清0后可以统一操作如两个长短不相同的与长短相同程序统一不用在分类讨论
memset(num2,0,sizeof(num2)) ;
memset(c,0,sizeof(c));
for(i=0;a[i]!='\0';++i) //将数组倒序排列方便操作序号对不好时举个例子如i=0
{
num1[i]=a[la-1-i]-'0';
}
for(i=0;b[i]!='\0';++i)
{
num2[i]=b[lb-1-i]-'0';
}

l=max(la,lb);
cnt=0;
for(i=0;i<=l ;++i) //主干程序模拟大数相加将其分解为一位数字的相加与进位两步主要利用循环思想就是找规律
{
c[i]=num1[i]+num2[i]+cnt;
if(c[i]>9)
{
c[i]-=10;
cnt=1;
}
else
{
cnt=0;
}
}
while(c[i]==0)
{
--i;
}

++k;
printf("Case %d:\n",k);
printf("%s + %s = ",a,b);
for(j=i;j>=0;--j)
{
printf("%d",c[j]);
}
if(t>0)
printf("\n"); //格式应当注意想好之后再提交不能应为小问题失掉比赛
printf("\n");
}
return 0;
}

基础练习（一）

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