hdu-1002 A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 355627    Accepted Submission(s): 68970

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

2

1 2 1

12233445566778899 998877665544332211

 

Sample Output

Case 1: 1 + 2 = 3

Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110

 

大数加法

我们都知道我们最常用的两种整数类型储存范围：

int ：[-2147483648,2147483647(2^31 - 1)] 我喜欢简单的记为如果超过9位数则不适合用int；

long long：[-9223372036854775808,9223372036854775807(2^63 - 1)]我喜欢简单地记为如果超过19位不适合用long long；

至于为什么我只说明了这两个，是因为我只习惯用这两个=。=；

如果Longlong都不合适，那么直接按照大数处理了

大数加法的方法我们可以看为小学加法竖式的思想来写，先用将两个大数存入两个字符串中，然后将两个字符串进行反转，存入两个数组中，然后从最末尾开始相加，然后将和反转，顺序输出即可，例如第二个样例：

112233445566778899和998877665544332211翻转后变成

998877665544332211

112233445566778899　+

-------------------------------

0111111111111111111

然后将得到的结果进行翻转，输出1111111111111111110

贴上代码：

#include<stdio.h>
#include<string.h>
int main()
{
    char s1[1010],s2[1010];
    int i,j,k=1,m,n,t;
    scanf("%d",&t);
    while(t--)
    {
        int a[1010]={0},b[1010]={0};
        scanf("%s%s",s1,s2);
        m=strlen(s1);n=strlen(s2);
        strrev(s1);strrev(s2);
        for(i=0;s1[i]!='';i++)
        a[i]=s1[i]-'0';
        for(i=0;s2[i]!='';i++)
        b[i]=s2[i]-'0';
        if(n>m)
        m=n;
        for(i=0;i<m;i++)
        {
            a[i]=a[i]+b[i];
            if(a[i]>9)
            {
                a[i+1]++;
                a[i]=a[i]-10;
            }
        }
        strrev(s1);strrev(s2);//太早之前写的代码，后来发现很多OJ不支持这个翻转函数，慎用。。。
        printf("Case %d:
",k);
        printf("%s",s1);
        printf(" + ");
        printf("%s",s2);
        printf(" = ");
        while(a[m]==0)
        m--;
        for(i=m;i>0;i--)
        printf("%d",a[i]);
        printf("%d",a[0]);
        printf("
");
        if(t!=0)
        printf("
");
        k++; 
    }
    return 0;
}

本来我想把这道题归到字符串里面，后来想了想，还是太鱼了~~

曾经我做这道题的时候还是一条鱼摆摆~