【HDU】5047 Sawtooth 高精度

传送门：【HDU】5047 Sawtooth

题目分析：公式还是蛮好推的，每增加一个‘M’就增加4条边，每条边和之前i个‘M’相交得到4*i个空间，再加上两个‘M’之间又形成一个空间，则增加第i+1个‘M’边增加了i*4+1个空间，最后可得公式f(n)=8*n*n-7*n+1。

这样算不算完了呢？显然不是，因为n高达1e12，long long也存不下，解决方法很多，如果用java则需要用读入优化，否则超时，我则用了高精度解决。

代码如下：

#include 
#include 
#include 
using namespace std ;

typedef long long LL ;

#define rep( i , a , b ) for ( int i = a ; i < b ; ++ i )
#define For( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define rev( i , a , b ) for ( int i = a ; i >= b ; -- i )

const int M = 10000 ;

struct BigInt {
	int digit[10] ;
	int length ;
	
	BigInt () {
		length = 0 ;
		memset ( digit , 0 , sizeof digit ) ;
	}
	
	BigInt ( LL number ) {
		length = 0 ;
		memset ( digit , 0 , sizeof digit ) ;
		while ( number ) {
			digit[length ++] = number % M ;
			number /= M ;
		}
	}
	
	int operator [] ( const int index ) const {
		return digit[index] ;
	}
	
	int& operator [] ( const int index ) {
		return digit[index] ;
	}
	
	BigInt fix () {
		while ( length && digit[length - 1] == 0 ) -- length ;
		return *this ;
	}
	
	BigInt operator + ( const BigInt& a ) const {
		BigInt c ;
		c.length = max ( length , a.length ) + 1 ;
		int add = 0 ;
		rep ( i , 0 , c.length ) {
			add += a[i] + digit[i] ;
			c[i] = add % M ;
			add /= M ;
		}
		return c.fix () ;
	}
	
	BigInt operator - ( const BigInt& a ) const {
		BigInt c ;
		c.length = max ( a.length , length ) ;
		int del = 0 ;
		rep ( i , 0 , c.length ) {
			del += digit[i] - a[i] ;
			c[i] = del ;
			del = 0 ;
			if ( c[i] < 0 ) {
				int tmp = ( c[i] - 1 ) / M + 1 ;
				c[i] += tmp * M ;
				del -= tmp ;
			}
		}
		return c.fix () ;
	}
	
	BigInt operator * ( const BigInt& a ) const {
		BigInt c ;
		c.length = a.length + length ;
		rep ( i , 0 , length ) {
			int mul = 0 ;
			For ( j , 0 , a.length ) {
				mul += digit[i] * a[j] + c[i + j] ;
				c[i + j] = mul % M ;
				mul /= M ;
			}
		}
		return c.fix () ;
	}
	
	void show () {
		printf ( "%d" , digit[length - 1] ) ;
		rev ( i , length - 2 , 0 ) printf ( "%04d" , digit[i] ) ;
		printf ( "\n" ) ;
	}
	
} ;

LL n ;

void scanf ( LL& x , char c = 0 ) {
	while ( ( c = getchar () ) < '0' || c > '9' ) ;
	x = c - '0' ;
	while ( ( c = getchar () ) >= '0' && c <= '9' ) x = x * 10 + c - '0' ;
}

void solve () {
	scanf ( n ) ;
	if ( n <= 1000000000 ) printf ( "%I64d\n" , 8 * n * n - 7 * n + 1 ) ;
	else {
		BigInt a = n ;
		a = a * ( a * 8 - 7 ) + 1 ;
		a.show () ;
	}
}		

int main () {
	int T , cas = 0 ;
	scanf ( "%d" , &T ) ;
	while ( T -- ) {
		printf ( "Case #%d: " , ++ cas ) ;
		solve () ;
	}
	return 0 ;
}