HDU 1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 204922    Accepted Submission(s): 39391

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
思路：1000位整数，用long long也存不下来。那么我们可以考虑用数组来存下这个数，再考虑一下进位跟输出格式，那么问题就简单了！
C++代码：
#include<stdio.h>#include<string.h>#define MAXN 1005int main(){    char a[MAXN],b[MAXN];    int a1[MAXN],a2[MAXN],c[MAXN];    int al,bl,k,i,cas,n,val,flag;    scanf("%d",&n);    getchar();    for(cas=1;cas<=n;cas++)    {        memset(a1,0,sizeof(a1));        memset(a2,0,sizeof(a2));        memset(c,0,sizeof(c));        if(cas!=1)            printf("\n");            scanf("%s %s",a,b);        al=strlen(a);        bl=strlen(b);        k=al>bl?al:bl;        for(i=0;i<al;i++)            a1[i]=a[al-i-1]-'0';<span style="white-space:pre"></span>//将字符转化为数字存入整形数组        for(i=0;i<bl;i++)<span style="white-space:pre"></span>//从低位到高位存            a2[i]=b[bl-i-1]-'0';        /*for(i=0;i<al;i++)            printf("%d",a1[i]);*/        val=0;        for(i=0;i<k;i++)        {            c[i]=val+a1[i]+a2[i];<span style="white-space:pre"></span>//val为进位数            val=c[i]/10;<span style="white-space:pre"></span>//进位            c[i]%=10;<span style="white-space:pre"></span>//进位后取整        }        printf("Case %d:\n",cas);         printf("%s + %s = ",a,b);        flag=0;        for(i=k-1;i>=0;i--)        {            if(c[i]!=0&&flag==0)//排除前导0<span style="white-space:pre"></span>                flag=1;            if(flag)<span style="white-space:pre"></span><span style="font-family: 'Courier New', Courier, monospace;">//从后面往前面输出</span>
<span style="font-family: 'Courier New', Courier, monospace; font-size: 12px;"></span>
<span style="font-family: 'Courier New', Courier, monospace; font-size: 12px;"></span>
<span style="font-family: 'Courier New', Courier, monospace; font-size: 12px;">              <span style="white-space:pre"></span>printf("%d",c[i]);</span>
                        }        printf("\n");     }    return 0;}
JAVA代码：
import java.math.BigInteger;import java.util.Scanner;public class Main{public static void main(String[] args) {Scanner cin= new Scanner(System.in);BigInterger a,b,c;int t=cin.nextInt();for(int cas=1;cas<=t;++cas){a=cin.nextBigInteger();b=cin.nextBigInteger();c=a.add(b);System.out.printIn("Case "+cas+":");System.out.printIn(a+" + "+b+" = "+c);if(cas<t)System.out.printIn();}System.out.println(bigInt);}}

感言：今天学好了大数相加并且能够初步地使用JAVA处理简单的问题，不错，再好好努力！