hduoj 1002 A + B Problem II（大数加法）

hduoj 1002 A + B Problem II（大数加法）

A + B Problem II

Time Limit: 2000/1000 MS (java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 301097    Accepted Submission(s): 58056

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

Author

Ignatius.L

AC：java大数

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

AC: C大数加法

#include#includeint main(){ char a[1000],b[1000],c[1001]; int i,j=1,p=0,n,n1,n2; scanf("%d",&n); while(n) { scanf("%s %s",a,b); printf("Case %d:\n",j); printf("%s + %s = ",a,b); n1=strlen(a)-1; n2=strlen(b)-1; for(i=0;n1>=0||n2>=0;i++,n1--,n2--) { if(n1>=0&&n2>=0){ c[i]=a[n1]+b[n2]-'0'+p;} if(n1>=0&&n2<0){ c[i]=a[n1]+p;} if(n1<0&&n2>=0){ c[i]=b[n2]+p;} p=0; if(c[i]>'9'){ c[i]=c[i]-10;p=1;} } if(p==1){printf("%d",p);} while(i--) { printf("%c",c[i]);} j++; if(n!=1){ printf("\n\n");} else {printf("\n");} n--; }}

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