【HDU】4927 Series 1 高精度

最新推荐文章于 2021-03-08 10:39:05 发布

poursoul

最新推荐文章于 2021-03-08 10:39:05 发布

阅读量913

点赞数

CC 4.0 BY-SA版权

分类专栏：高精度【Bignumber】文章标签： HDU

本文链接：https://blog.youkuaiyun.com/u013368721/article/details/38423871

高精度【Bignumber】专栏收录该内容

4 篇文章

订阅专栏

本文介绍了 HDU 4927 Series1 的解题思路及大数模板实现方法。通过组合数学推导出公式，并使用自定义大数类进行计算，解决了整数溢出的问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

传送门：【HDU】4927 Series 1

题目分析：公式很好推，到最后就是C(n-1,0)*a[n]-C(n-1,1)*a[n-1]+C(n-1,2)*a[n-2]+...+C(n-1,n-1)*a[n]。

用C(n,k)=C(n,k-1)*(n-k+1)/k即可快速得到一行的二项式系数。

我看JAVA不到1000B 15分钟就能过。。。我又敲了大数模板然后将近2个小时才过T U T......

不过大数模板敲起来还是蛮爽的。。。就是暂时不能实现大数除法以及带负数的运算（仅限非负整数）

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

#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 )
#define CLR( a , x ) memset ( a , x , sizeof a )
#define CPY( a , x ) memcpy ( a , x , sizeof a )

const int L = 10000 ;
const int MAXN = 300 ;

struct BigInt {
	int length , digit[MAXN] ;
	
	BigInt ( int number = 0 ) {
		CLR ( digit , 0 ) ;
		length = 0 ;
		while ( number ) {
			digit[length ++] = number % L ;
			number /= L ;
		}
	}
	
	BigInt fix () {
		while ( length && !digit[length - 1] )
			-- length ;
		return *this ;
	}
	
	BigInt operator = ( int number ) {
		CLR ( digit , 0 ) ;
		length = 0 ;
		while ( number ) {
			digit[length ++] = number % L ;
			number /= L ;
		}
		return *this ;
	}
	
	int operator [] ( const int &index ) const {
		return digit[index] ;
	}
	
	int & operator [] ( const int &index ) {
		return digit[index] ;
	}
	
	BigInt operator + ( const BigInt &b ) const {
		BigInt c ;
		c.length = max ( length , b.length ) + 1 ;
		int add = 0 ;
		REP ( i , 0 , c.length ) {
			add += digit[i] + b[i] ;
			c[i] = add % L ;
			add /= L ;
		}
		return c.fix () ;
	}
	
	BigInt operator - ( const BigInt &b ) const {
		BigInt c ;
		c.length = max ( length , b.length ) ;
		int del = 0 ;
		REP ( i , 0 , c.length ) {
			del += digit[i] - b[i] ;
			c[i] = del ;
			del = 0 ;
			if ( c[i] < 0 ) {
				int tmp = ( -c[i] - 1 ) / L + 1 ;
				c[i] += tmp * L ;
				del -= tmp ;
			}
		}
		return c.fix () ;
	}
	
	BigInt operator * ( const BigInt &b ) const {
		BigInt c ;
		c.length = length + b.length ;
		REP ( i , 0 , length ) {
			int mul = 0 ;
			FOR ( j , 0 , b.length ) {
				mul += digit[i] * b[j] + c[i + j] ;
				c[i + j] = mul % L ;
				mul /= L ;
			}
		}
		return c.fix () ;
	}
	
	BigInt operator / ( const int &b ) const {
		BigInt c ;
		c.length = length ;
		int over = 0 ;
		REV ( i , length - 1 , 0 ) {
			over = over * L + digit[i] ;
			c[i] = over / b ;
			over %= b ;
		}
		return c.fix () ;
	}
	
	BigInt operator += ( const BigInt &b ) {
		*this = *this + b ;
		return *this ;
	}
	
	BigInt operator -= ( const BigInt &b ) {
		*this = *this - b ;
		return *this ;
	}
	
	BigInt operator *= ( const BigInt &b ) {
		*this = *this * b ;
		return *this ;
	}
	
	BigInt operator /= ( const int &b ) {
		*this = *this / b ;
		return *this ;
	}
	
	bool operator < ( const BigInt &b ) const {
		if ( length != b.length )
			return length < b.length ;
		REV ( i , length - 1 , 0 )
			if ( digit[i] != b[i] )
				return digit[i] < b[i] ;
		return false ;
	}
	
	bool operator > ( const BigInt &b ) const {
		return b < *this ;
	}
	
	bool operator <= ( const BigInt &b ) const {
		return !( b < *this ) ;
	}
	
	bool operator >= ( const BigInt &b ) const {
		return !( *this < b ) ;
	}
	
	bool operator != ( const BigInt &b ) const {
		return b < *this || *this < b ;
	}
	
	bool operator == ( const BigInt &b ) const {
		return !( b < *this ) && !( *this < b ) ;
	}
} ;

int A[L] ;

void print ( const BigInt &res ) {
	printf ( "%d" , res[res.length - 1] ) ;
	REV ( i , res.length - 2 , 0 )
		printf ( "%04d" , res[i] ) ;
	printf ( "\n" ) ;
}

void solve () {
	int n , nn ;
	BigInt res = 0 , positive = 0 , negative = 0 , Cij = 1 ;
	scanf ( "%d" , &n ) ;
	REV ( i , n , 1 )
		scanf ( "%d" , A + i ) ;
	FOR ( i , 1 , n ) {
		if ( i > 1 )
			Cij = Cij * ( n - i + 1 ) / ( i - 1 ) ;
		if ( i & 1 )
			positive += Cij * A[i] ;
		else
			negative += Cij * A[i] ;
	}
	//print ( positive ) ;
	//print ( negative ) ;
	if ( positive < negative ) {
		printf ( "-" ) ;
		res = negative - positive ;
	}
	else
		res = positive - negative ;
	print ( res ) ;
}

int main () {
	int T ;
	scanf ( "%d" , &T ) ;
	while ( T -- )
		solve () ;
	return 0 ;
}