HDU 1002 大数

发布时间：2020-12-14 04:03:27 所属栏目：大数据来源：网络整理

导读：? 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

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 = 2222222222222221110

#include <iostream>
using namespace std;
int main()
{
	char s1[1111],s2[1111];
	int n1,n2,n,A[1111],t,j;
	scanf("%d",&t);
	for(j=1;j<=t;j++)
	{
		if(j>1)printf("n");
		scanf("%s%s",&s1,&s2);
		n1=strlen(s1),n2=strlen(s2);
		n=n1>n2?n1:n2;
		memset(A,sizeof(A));
		int k=0,i;
		for(i=0;i<n;i++)//个位数相加
		{
			n1--;n2--;
			if(n1>=0&&n2>=0)
				A[k++]=s1[n1]+s2[n2]-2*'0';
			else if(n1>=0&&n2<0)
				A[k++]=s1[n1]-'0';
			else if(n1<0&&n2>=0)
				A[k++]=s2[n2]-'0';
		}
		for(i=0;i<k-1;i++)//进位
			if(A[i]>9)A[i+1]++,A[i]%=10;
		for(i=k-1;i>0;i--)//去除前面为0的
			if(A[i])break;
		k=i; 
		printf("Case %d:n%s + %s = ",j,s1,s2);
		for(i=k;i>=0;i--)printf("%d",A[i]);
		printf("n");
	}
	return 0;
}

以前做的：

#include<stdio.h>
#include<string.h>
int main()
{ 
    char a[1001],b[1001];      
    int  i,k=0;
    scanf("%d",&t);
    while(t--)
    { 
        if(k>0)
            puts("");
        k++;
        int A[1001]={0},B[1001]={0};
        scanf("%s%s",a,b);
        int x=strlen(a),y=strlen(b);
        for(i=0;i<x;i++)
            A[i]=a[x-1-i]-'0';
        for(i=0;i<y;i++)
            B[i]=b[y-1-i]-'0';
        int c=0;
        for(i=0;i<1001;i++)
        {  
            int s=(A[i]+B[i]+c);
            A[i]=s%10;
            c=s/10;
        }
        printf("Case %d:n%s + %s = ",k,b);
        for( i=1001-1;i>=0;i--) 
            if(A[i])
                break;
        for(j=i;j>=0;j--)     
            printf("%d",A[j]);
        puts("");
    }
    return 0;
}

（编辑：李大同）

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