HDOJ Problem 1002 - A + B Problem II

HDOJ Problem 1002 - A + B Problem II:

大数定理

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

1
2
3

2
1 2
112233445566778899 998877665544332211

Sample Output

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

解法一:

#include <iostream>
#include <string.h>
using namespace std;
char a[1000],b[1000];
int aar[1001],bar[1001];

int main()
{
	int alen,blen,maxlen;
	int time;
	cin >> time;
	for (int i=1;i<=time;i++)
	{
		cin >> a >> b;
		alen = strlen(a);
		blen = strlen(b);
		int tmp = 0;
		for (int j=alen-1;j>=0;j--)
			aar[tmp++] = a[j]-'0';
		tmp = 0;
		for (int j=blen-1;j>=0;j--)
			bar[tmp++] = b[j]-'0';
		//123 ==>  321 为了保证之后便于计算
		
		if ( alen > blen){
			maxlen = alen;
			for(int j=blen;j<alen;j++){	//长度不同时,短的那个需要补零
				bar[j] = 0;
			}
			aar[alen] = 0;
		}
		else if ( alen < blen){
			maxlen = blen;
			for(int j=alen;j<blen;j++)
			{
				aar[j] = 0;
			}
			bar[blen] = 0;
		}
		else{
			maxlen = alen;
			aar[maxlen]=0;
			bar[maxlen]=0;
		}
		
		for (int j=0;j<maxlen;j++){
			cout << aar[j] << '\t';
			cout << bar[j] <<endl;
			aar[j] += bar[j];
			
			if(aar[j] >= 10){//如果当前位大于10则进一位
				aar[j] -=10;
				aar[j+1] += 1; 
			}
		}//将a+b的和保存在aar数组里
		
		cout << "Case:" << i<< endl;
		cout << a << " + " << b << " = " ;
		
		if (aar[maxlen] == 0 )		//判断第一位是否为0,如果是的话就从第二位开始读,这个是逆序
			for (int j=maxlen-1;j>=0;j--) cout << aar[j];
		else
			for (int j=maxlen;j>=0;j--)
				cout << aar[j];

		if(i != time)			//注意要求的输出格式
			cout << endl<< endl;
		else
			cout << endl;
	}
	return 0;
}

解法二:

#include<iostream>
#include<string>
#include<cstdio>
using namespace std;

int i,j,y,n,k,h,p,lena,lenb;
int a[1000],b[1000],sum[1000],f[1000];
int main()
{
	string a1,b1;
	cin>>n;
	for(y=1;y<=n;y++)
	{
		cin>>a1>>b1;
		lena=a1.length();
		lenb=b1.length();
		for(i=0;i<1000;i++){
			a[i]=0;b[i]=0;f[i]=0;//f数组是记录a+b和		
        //这个补零是先全都设为0,再把不为0的填入
		}
		for(i=lena-1;i>=0;i--)		/*1234 ==> 4321*/
		    a[lena-1-i]=a1[i]-'0';	//字符'9' - '0' 才是数字9
		for(i=lenb-1;i>=0;i--)
		    b[lenb-1-i]=b1[i]-'0';
	    k=0;
		for(i=0;i<lenb || i<lena;i++){		//i--> max( lena , lenb )
			h=a[i]+b[i]+k;	//k是下一位是否进一
			f[i]=h%10;		//f[i]必然是0-9
			k=h/10;			//如果h大于10,则k=1,如果h小于10,则k=0
		}
		if(k!=0)			//如果k=1,则最高位加一
		    f[i++]=k; 
		cout<<"Case "<<y<<":"<<endl<< a1 <<" + "<< b1 <<" = ";
		p=0;
		for(j=i-1;j>=0;j--){			//将之前为了计算时的倒序,再反正过来
		 	if(p==0 && f[j]==0){		//目的是去前导0,实则这步多余了
		 	    continue;
                //如果最高位是0的话i就不会++,如果是1的话,那么f[j]就不会是0,所以这步必然是进入else
		 	}
		 	else{
		 	    p=1;
		 	    cout<<f[j];
		 	}
		}
		cout<<endl;
	}
	return 0;
}

▲总结:

1.为什么用字符数组==>因为数字太大,long long也存储不下
2.用int数组记录每一位的数字,然后模拟手算
3.为什么倒置==>因为为了让末尾对齐,方便计算"个位对个位,十位对十位……"