hdu 1002 A + B Problem II（大数相加）

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A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 362459    Accepted Submission(s): 70479


[align=left]Problem Description[/align]
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
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.

 

[align=left]Sample Input[/align]

2
1 2
112233445566778899 998877665544332211

 

[align=left]Sample Output[/align]

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

 

[align=left]Author[/align]
Ignatius.L
 
本题题意：先输入一个数 t，表示有 t 组测试数据库，然后给出两个数，注意的是这两个数很大，甚至超出长整型的范围，然后计算这两个数的和，输出格式是先输出Case #:其中 ‘#’ 表示第几组测试数据，然后接下来一行输出结果。每组测试数据之后再空一行，最后一组末尾没有空行。

解题思路：大数相加，一般都是用数组先存起来，然后分别取数组最后一位相加，若相加大于10，就用p值矫正进位，若其中一个数组用完，就直接记录该值就好，最后再倒序输出即可。

代码1：

#include <stdio.h>
#include <string.h>
int main()
{
char s1[1005],s2[1005], s[1005];//s数组存为和值
int x=1;
int t;
int len1,len2,k;
scanf("%d",&t);
while(t--)
{
int p=0;
scanf("%s %s",s1,s2);
printf("Case %d:\n",x++);
len1=strlen(s1)-1;    //第一个数的长度
len2=strlen(s2)-1;   //第二个数的长度
for(k=0; len1>=0|| len2>=0;  len1--, len2--,k++)   //当s1和s2数组长度均为0时退出循环，都从最后一位开始相加
{
if( len1>=0&& len2>=0) s[k]=s1[len1]+s2[ len2]-'0'+p;
if( len1<0&& len2>=0)  s[k]=s2[len2]+p;     //第一个数据
if( len1>=0&& len2<0)  s[k]=s1[len1]+p;
p=0;                                     //记得还原
if(s[k]>'9')
{
s[k]=s[k]-10;
p=1;      //该位相加大于十就进位
}
}
printf("%s + %s = ",s1,s2);
if(p==1)             //最后的数大于十就先输出进位1
printf("1");
while(k--)          //倒序输出
printf("%c",s[k]);
if(t!=0)
printf("\n\n");
else
printf("\n");
}
return 0;
}

代码2(分步计算):

#include <stdio.h>
#include <stdlib.h>
#include<string.h>
int main()
{
int t,x=1,i,j,k,lena,lenb,p;
char a[20000],b[20000];
scanf("%d",&n);
while(t--)
{
int s[20000]= {0};
printf("Case %d:\n",x++);
scanf("%s%s",a,b);
lena=strlen(a);
lenb=strlen(b);
printf("%s + %s = ",a,b);
j=lena-1;
k=lenb-1;
p=0;
while(j>=0&&k>=0)  //当数组a和数组其中一个数组的长度为0时退出
{
if(s[p]+(a[j]-'0')+(b[k]-'0')>=10)  //进位
{
s[p]=s[p]+(a[j]-'0')+(b[k]-'0')-10;
s[p+1]++;//数组s的之后的一位加一
}
else
s[p]=s[p]+(a[j]-'0')+(b[k]-'0'); //否则直接相加
p++;
k--;
j--;
}
//执行完前面的代码之后必定只剩下一个数组的值了
if(j>=0)
{
for(i=j; i>=0; i--)
{
s[p]=s[p]+(a[i]-'0');
p++;
}
}
else if(k>=0)
{
for(i=k; i>=0; i--)
{
s[p]=s[p]+b[i]-'0';
p++;
}
}
while(p--)
printf("%d",s[p]);//倒序输出
printf("\n");
if(t!=0)
printf("\n");
}
return 0;
}