杭电ACM1002--大数相加

最新推荐文章于 2018-08-07 21:57:11 发布

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本文介绍了一种解决大数加法问题的方法，通过使用Java的BigInteger类和C/C++的数组实现，解决了传统32位整数溢出的问题。适用于长度不超过1000的大数相加。

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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

2
1 2
112233445566778899 998877665544332211

Sample Output

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110


1、用java的BigInteger类来做非常简单.

import java.math.BigInteger;
import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t;
        BigInteger n,m;
        t=sc.nextInt();
        for(int i=0 ;  i<t ; i++){
            if(i>0)System.out.println();
            n = new BigInteger(""+sc.nextBigInteger());
            m = new BigInteger(""+sc.nextBigInteger());
            System.out.println("Case "+(i+1)+":");
            System.out.println(n+" + "+m+" = "+m.add(n));
        }
    }
}

2、用c/c++来写代码相对较长，用数组来做

#include<stdio.h>
#include<string.h>
int main ()
{
    int a[1002],b[1002],Case=0,t,i,l,j,p;
    char x[1002],y[1002];
    scanf("%d",&t);
    while(t--)
    {
        getchar();
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        scanf("%s%s",x,y);
        l=strlen(x)>strlen(y)?strlen(x):strlen(y);
        j=p=0;
        for(i=l-1; i>=0; i--)
        {
            if(x[i]>='0'&&x[i]<='9')
            {
                a[j]=x[i]-'0';
                j++;
            }
            if(y[i]>='0'&&y[i]<='9')
            {
                b[p]=y[i]-'0';
                p++;
            }
        }
        printf("Case %d:\n",++Case);
        for(i=0; i<l; i++)
        {
            if(a[i]+b[i]>=10)
            {
                a[i]=(a[i]+b[i])-10;
                a[i+1]++;
            }
            else
            {
                a[i]=a[i]+b[i];
            }
        }
        printf("%s + %s = ",x,y);
        if(a[l]!=0)printf("%d",a[l]);
        for(i=l-1; i>=0; i--)
        {
            printf("%d",a[i]);
        }
        if(t==0)printf("\n");
        else printf ("\n\n");
    }
    return 0;
}