两个无穷大的数相加的问题

[yeshengyu]

这是原题, 哪位高手能知道一下啊?



A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11377 Accepted Submission(s): 1904


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

[ryfdizuo]

用int64_就可以了，baidu一下

[星羽]

google 大数运算

[ryfdizuo]

#include<stdio.h>
#include<string.h>
int main()
{
	char str1[1000],str2[1000];     
	int a[1000],b[1000],c[1001];
	int N;
	int m,n,i,j,s,x,k,h,t;
	int flag=0;
	int len1,len2,len;

	scanf("%d",&N);
	for(i=1;i<=N;++i)
	{
		scanf("%s %s",&str1,&str2);
		printf("Case ");
		printf("%d",i);
		printf(":\n");
		printf("%s + %s = ",str1,str2);

		len1=strlen(str1); 
		len2=strlen(str2);

		for(s=len1-1,m=0;s>=0;s--)
		{
			a[m]=str1[s]-'0';
			m++;
		}
		for(j=len2-1,n=0;j>=0;j--)
		{
			b[n]=str2[j]-'0';
			n++;
		}
		for(h=len1;h<1000;h++)
			a[h]=0;
		for(t=len2;t<1000;t++)
			b[t]=0;
		if(len1>len2)
			len=len1;
		else if(len1=len2)
			len=len1;
		else
			len=len2;
		for(s=0;s<len;s++)
		{
			x=a[s]+b[s]+flag;
			if(x>=10)
			{
				c[s]=x%10;
				flag=1;
			}
			else
			{
				c[s]=x;
				flag=0;
			}
		}
		if(flag==1)
			printf("1");
		for(k=len-1;k>=0;k--)
			printf("%d",c[k]);    
		printf("\n");
		if (i<N)
			printf("\n");
	}
	return 0;
}

[kilvdn]

#include <stdio.h> 
#include <string.h> 

int main(){ 
    char a[1005],b[1005]; 
    int kase,ase=1;;
    scanf("%d",&kase);
    while (kase--){
        
//    while (scanf("%s",a)!=EOF){ 
        scanf ("%s%s",a,b); 
        if (ase>1)  printf("\n");
        printf("Case %d:\n",ase++);
        
        int la,lb; 
        la=strlen(a); 
        lb=strlen(b); 
        printf("%s + %s = ",a,b);
        char c[2020]; 
        int i,j,k=0,jw=0; 
        for (i=la-1,j=lb-1;i>=0&&j>=0;i--,j--){ 
            c[k++]=(a[i]-'0'+b[j]-'0'+jw)%10+'0'; 
            jw=(a[i]-'0'+b[j]-'0'+jw)/10; 
        } 
        if (i==-1){ 
            for (j;j>=0;j--){ 
                c[k++]=(b[j]-'0'+jw)%10+'0'; 
                jw=(b[j]-'0'+jw)/10; 
            } 
        } 
        else { 
            for (i;i>=0;i--){ 
                c[k++]=(a[i]-'0'+jw)%10+'0'; 
                jw=(a[i]-'0'+jw)/10; 
            } 
        } 
        if(jw!=0) 
        printf ("%d",jw); 
        for (k-=1;k>=0;k--) 
        printf ("%d",c[k]-'0'); 
        printf("\n"); 
    } 
return 0 ;  
} 

[bm1408]

上学的时候也做过,给你点提示,采用链表方式这实现

[giftfish]

用数组

[yeshengyu]

啊?你用的是JAVA啊,我学的是C语言,呵呵

[freeCodeSunny]

import java.io.*;
import java.util.*;
import java.math.BigInteger;
public class A{
public static void main(String[] args){
BigInteger sum=new BigInteger("0");
BigInteger num1,num2;
Scanner cin=new Scanner(System.in);
num1=cin.nextBigInteger();
num2=cin.nextBigInteger();
sum=sum.add(num1);
sum=sum.add(num2);
System.out.println(sum.toString());

}
}

[freeCodeSunny]

上面要把A改成Main

[昨日凡阳]

http://blog.csdn.net/fenglibing/article/details/1756773