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  • A + B Problem II(杭电1002)

    /*A + B Problem II
    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*/

    /*借鉴大神经验。与新手共勉。


     

    */



    #include<stdio.h>
    #include<string.h>
    int main()
    {
        int str1[1100],str2[1100];
        char a[1100],b[1100];
        int test,i,j,t,k=1,len1,len2;
        scanf("%d",&test);
        getchar();
        while(test--)
        {
            char sum[1100]={0};
            scanf("%s %s",a,b);
            getchar();
            len1=strlen(a);
            len2=strlen(b);
            memset(str1,0,sizeof(str1));
            memset(str2,0,sizeof(str2));
            for(i=0,j=len1-1;i<len1;i++,j--)
            str1[j]=a[i]-'0';
            for(i=0,j=len2-1;i<len2;i++,j--)
            str2[j]=b[i]-'0';
            if(len1<len2)
            {
                t=len2;
                len2=len1;
                len1=t;
            }
            for(i=0;i<=len1;i++)
            {
                sum[i]=str1[i]+str2[i]+sum[i];
                if(sum[i]/10>0)
                {
                    sum[i+1]=sum[i]/10;
                    sum[i]=sum[i]%10;
                }
            }
            printf("Case %d:
    ",k);
            k++;
            printf("%s + %s = ",a,b);
            if(sum[len1]!=0)
            printf("%d",sum[len1]);
            for(j=len1-1;j>=0;j--)
            printf("%d",sum[j]);
            printf("
    ");
            if(test!=0)
            printf("
    ");
        }
        return 0;
    }


    今天尝试用java写这道大数题目,贴下代码:


    import java.math.BigInteger;
    import java.util.Scanner;
    public class Main {


    public static void main(String[] args) {

    Scanner in=new Scanner(System.in);
    int test=in.nextInt();
    String s1,s2;
    int kase=1;
    while(test-->0)
    {
    s1=in.next();
    s2=in.next();
    BigInteger s3=new BigInteger(s1);
    BigInteger s4=new BigInteger(s2);
    System.out.println("Case "+kase+++":");
    System.out.println(s3+" + "+s4+" = "+s3.add(s4));
    if(test!=0)
    System.out.println();
    }
    }


    }

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  • 原文地址:https://www.cnblogs.com/wzjhoutai/p/7152751.html
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