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): 122234    Accepted Submission(s): 23410

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

 

 

 小学加法怎么做，程序就怎么编。

AC代码：

#include<stdio.h>#include<string.h>int shu(char a){    return (a-'0');}int main(){    char a[1000],b[1000];    int num[1001];    int n,i,j=1,al,bl,k,t;    scanf("%d",&n);    while(n--)    {       getchar();        if(j!=1)       printf("\n");       scanf("%s",a);       al=strlen(a);       scanf("%s",b);       bl=strlen(b);       k=(al>bl)?al:bl;       for(i=0;i<=k;i++)       num[i]=0;       t=k;       for(k;al>0&&bl>0;k--)       {           num[k]+=shu(a[--al])+shu(b[--bl]);           if(num[k]/10)           {               num[k-1]++;               num[k]%=10;           }       }       while(al>0)       {            num[k--]+=shu(a[--al]);            if(num[k+1]/10)           {               num[k]++;               num[k+1]%=10;           }       }       while(bl>0)       {            num[k--]+=shu(b[--bl]);            if(num[k+1]/10)           {               num[k]++;               num[k+1]%=10;           }       }       printf("Case %d:\n",j++);       printf("%s + %s = ",a,b);       for(i=0;i<=t;i++)       {           if(i==0&&num[i]==0)           i++;           printf("%d",num[i]);       }       printf("\n");   }   return 0;}

 最优解：

 import java.math.BigInteger;import java.util.Scanner;public class Main{    public static void main(String args[]) {       Scanner cin=new Scanner(System.in);       int n=cin.nextInt();       BigInteger a,b;       for(int i=1;i<=n;i++){       a=cin.nextBigInteger();       b=cin.nextBigInteger();       System.out.println("Case "+i+":");       System.out.println(a.toString()+" + "+b.toString()+" = "+a.add(b));       }    }}