HDU 1002 A + B Problem II（ 高精度加法水 ）

---

**链接：****传送门 **

题意：A + B 高精度，板子题

---

 /*************************************************************************
    > File Name: hdu1002.cpp
    > Author:    WArobot 
    > Blog:      http://www.cnblogs.com/WArobot/ 
    > Created Time: 2017年05月16日 星期二 20时08分28秒
 ************************************************************************/

#include<bits/stdc++.h>
using namespace std;

#define cls(x) memset(x,0,sizeof(x))

const int maxlen = 10000;	// 单个“数字”最大长度
class HP{		// High Precision
public:
	int len , s[maxlen];	
	HP(){ (*this) = 0; };	HP(int inte){ (*this) = inte; };	HP(const char*str){ (*this) = str; };
	friend ostream & operator << (ostream & cout , const HP &x);		// 定义高精度输出方式
	HP operator = (int inte);		HP operator = (const char*str);		// 定义高精度与高精度的 + - * / % Compare(比较)
	HP operator	+ (const HP &b);	HP operator - (const HP &b);
	HP operator * (const HP &b);	HP operator / (const HP &b);
	HP operator % (const HP &b);	int Compare(const HP &b);
};
ostream & operator <<(ostream & cout, const HP &x){
	for(int i = x.len ; i >= 1 ; i--)	cout<< x.s[i];	return cout;
}
HP HP::operator = (int inte){
	if( inte == 0 ){ len = 1; s[1] = 0; return (*this); };
	for( len = 0 ; inte > 0 ;) { s[++len] = inte % 10 ; inte /= 10; };
	return (*this);
}
HP HP::operator = (const char* str){
	len = strlen(str);
	for(int i = 1 ; i <= len ; i++)	s[i] = str[len - i] - '0';	// 需要注意
	return (*this);
}
HP HP::operator * (const HP &b){
	int i , j;	HP c;	c.len = len + b.len;
	for( i = 1 ; i <= c.len ; i++)	c.s[i] = 0;			// 清空返回值的数组
	for( i = 1 ; i <= len ; i++)	for( j = 1 ; j <= b.len ; j++)	 c.s[i+j-1] += s[i]*b.s[j];
	for( i = 1 ; i < c.len ; i++){  c.s[i+1] += c.s[i] / 10 ; c.s[i] %= 10; }
	while( c.s[i] ){ c.s[i+1] = c.s[i]/10 ; c.s[i] %= 10; i++; }
	while( i>1 && !c.s[i] )	i--; c.len = i;
	return c;
}
HP HP::operator + (const HP &b){	
	int i;	HP c; c.s[1] = 0;
	for( i = 1 ; i<=len || i<=b.len || c.s[i] ; i++ ){	// 模拟十进制加法运算，写的好呀！写的好！巧妙！
		if( i <= len )		c.s[i] += s[i];
		if( i <= b.len )	c.s[i] += b.s[i];
		c.s[i+1] = c.s[i]/10;	c.s[i] %= 10;
	}
	c.len = i - 1 ;		if( c.len == 0 )	c.len = 1;
	return c;
}
HP HP::operator - (const HP &b){
	int i,j;	HP c;
	for( i = 1, j = 0; i <= len ; i++){
		c.s[i] = s[i] - j;	if(i <= b.len) c.s[i] -= b.s[i];
		if( c.s[i] < 0 ){ j = 1; c.s[i] += 10; }	
		else			  j = 0;
	}
	c.len = len;	while( c.len > 1 && !c.s[c.len] )	c.len--;
	return c;
}
int HP::Compare(const HP &y){
	if( len > y.len )	return 1;
	if( len < y.len )	return -1;
	int i = len;
	while( (i>1) && (s[i]==y.s[i]) ) i--;
	return s[i] - y.s[i];			// 如果 s[i] == y.s[i] 自然返回0,s[i] > y.s[i]返回1.....很巧妙！
}
HP HP::operator / (const HP &b){
	int i,j;	HP d(0) , c;
	for( i = len ; i > 0 ; i--){
		if( !(d.len==1 && d.s[1]==0) )
			{ for( j = d.len ; j > 0 ; j--) d.s[j+1] = d.s[j];	++d.len; }
		d.s[1] = s[i];	c.s[i] = 0;
		while( j = d.Compare(b) >= 0 ){
			d = d - b;	c.s[i]++;	if(j==0)	break;
		}
	}
	c.len = len ;	while( (c.len>1) && (c.s[c.len]==0) )	c.len--;
	return c;
}
HP HP::operator % (const HP &b){
	int i,j;	HP d(0);
	for( i = len ; i > 0 ; i--){
		if( !(d.len==1 && d.s[1]==0) )
			{ for( j = d.len ; j > 0 ; j--) d.s[j+1] = d.s[j];	++d.len; }
		d.s[1] = s[i];
		while( j = d.Compare(b) >= 0 ){
			d = d - b;		if(j==0) break;
		}
	}
	return d;
}
int main(){
	int t,kase = 0;
	HP a,b,c;
	char s1[maxlen] , s2[maxlen];
	scanf("%d",&t);
	while(t--){
		if( kase )	puts("");
		printf("Case %d:
",++kase);
		cls(s1);	cls(s2);
		scanf("%s %s",s1,s2);
	
		a = s1;		b = s2;
		c = a + b;
		cout<< a << " + " << b << " = " << c <<endl;
	}
	return 0;
}