HDU 1002

最新推荐文章于 2020-02-02 15:34:19 发布

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本文介绍了如何处理大数相加问题，并通过JAVA语言实现。重点在于理解大数处理逻辑，掌握基本的JAVA操作，包括使用BigInteger类进行大数运算。通过实例演示了从输入到输出的完整流程，包括读取输入数据、处理大数相加、格式化输出等关键步骤。

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

2 1 2 112233445566778899 998877665544332211

Sample Output

Case 1: 1 + 2 = 3 Case 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 1005
int 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！