杭電HDOJ 1002 解題報告

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 154489    Accepted Submission(s): 29230

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

大數問題。

解題思路：

既然int、double等都無法解決如此大的數，那麼我們可以用數組來解決。

PS：當然也可以考慮下__int64，是否能解決呢？它的範圍是多少呢？int和double的範圍是多少呢？Unsigned呢？

當然也要注意回車、數組大小、空格以及進位哦！

如果你還有興趣的話，你也可以考慮下100位的2進制數如何轉換成10進制數呢？100位的2進制數如何轉換成8進制數呢？

代碼：

#include<stdio.h>
#include<string.h>
int m,p=0,k=0;

void sum(char a[],char b[])
{
	char s;
	if(k<m){
		s=a[k]+b[k]-48+p;p=0;
		if((s-48)/10){
			s-=10;
			p=1;
		}k++;
		sum(a,b);
		printf("%c",s);
	}else if(p==1){printf("1");}
}

void reverse(char s[],int l)
{
	int i,j,temp;

	for(i=0,j=l-1;i<j;i++,j--)
	{temp=s[i];s[i]=s[j];s[j]=temp;}
}

main()
{
	char a[20][1000],b[20][1000];
	int l1,l2,i,j,n;

	scanf("%d",&n);
	for(i=0;i<=n-1;i++)
	{
        scanf("%s",&a[i]);scanf("%s",&b[i]);
		printf("Case %d:\n",i+1);
		printf("%s + %s = ",a[i],b[i]);
		l1=strlen(a[i]);l2=strlen(b[i]);
		m=l1>l2?l1:l2;
		reverse(a[i],l1);
		reverse(b[i],l2);
		if(l1>l2){
			for(j=l2;j<l1;j++)
				b[i][j]='0';
		}
		else if(l1<l2){			
			for(j=l1;j<l2;j++)
				a[i][j]='0';
		}
		sum(a[i],b[i]);
		p=0;k=0;
		printf("\n");
		if(i!=n-1) printf("\n");
	}
}