hdu 大数据相加-a+b Problem II

A + B Problem II

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 108   Accepted Submission(s) : 21

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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

 

此题的数据若为64位以内，那做起来就so easy，但它的数据远超过了64位，只好按数组处理了，另外还要注意输出格式。

算法：首先思考一下中学时计算多位数时的竖式算法，对应位相加，满十进一。此题就是用这种方法，把计算得来的每一位存入数组，大数相加便解决了。

输入包括多组数据，第一行输入处理的数据数n，接下来输入n行，每行输入两个数a和b.题目要求算出a+b,

输出格式为：

case空格x：

a空格+空格b空格=空格a+b

每组结果之间有一个空行最后一组没有

代码：

#include <iostream>#include <cstring>using namespace std;int main(){    char a[10000],b[10000];    int h,p,k,j=0,i=0,n,s,la,lb,j1,i1;    cin>>n;    for(s=1;s<=n;s++)    {     int s1[10001]={0},s2[10001]={0},s3[100000]={0};     cin>>a;//输入字符串a     la=strlen(a);//测量字符串的长度     //把字符串a转换成数组     for(i=la-1,i1=0;i>=0;i--,i1++)     {       s1[i1]=a[i]-48;     }     //字符串b同a处理     cin>>b;     lb=strlen(b);     for(j=lb-1,j1=0;j>=0;j1++,j--)     {         s2[j1]=b[j]-48;     }     p=((lb<la)? la:lb);//比较字符串a和b的长度把大数赋给p     h=0;     for(k=0;k<p;k++)     {        s3[k]=(s1[k] + s2[k] + h) % 10;        h= (s1[k] + s2[k] + h) / 10;     }     cout<<"Case"<<' '<<s<<":"<<endl;     cout<<a<<' '<<"+"<<' '<<b<<' '<<"="<<' ';     if(s3[k]!=0) cout<<'1';     for(j=k-1;j>=0;j--)     {       cout<<s3[j];     }     cout<<endl;     if(s<n)     cout<<endl;    }    return 0; }

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