BZOJ-序列统计
$$HZOI$$
·题意
给你 \(n\) , \(l\) , \(r\) , 你可以于 \([ l , r ]\) 中选 \(n\) , 个数 , 组成的序列为不下降序列的个数 , 对 \(mod = 10^9 + 7\)
\(n \le 10^9 , l \le r \le 10 ^ 9\)
·题解
明显的得到式子：
\[ans = \sum^{n}_{i = 1}C^{i}_{r - l + 1 + i}
\]
用杨辉三角化一下得到：
\[C^{n}_{n + r - l + 1} - 1
\]
code
#include<bits/stdc++.h>
#define int long long
#define lson ( id << 1 )
#define rson ( id << 1 | 1 )
using namespace std ;
const int N = 1000010 ;
const int mod = 1e6 + 3 ;
namespace Combination
{int D[ N ] ; int nueyong[ N ] , sum_neo[ N ] , sum[ N ] ; inline void lear_neoyong( ){sum_neo[ 0 ] = sum_neo[ 1 ] = 1 ; nueyong[ 1 ] = 1 ; nueyong[ 0 ] = 1 ; sum[ 0 ] = sum[ 1 ] = 1 ; for( int i = 2 ; i < N ; ++ i ){int p = mod ; int k = p / i ; nueyong[ i ] = ( k * ( p - nueyong[ p % i ] ) ) % p ; sum_neo[ i ] = ( nueyong[ i ] * sum_neo[ i - 1 ] ) % p ; sum[ i ] = ( i * sum[ i - 1 ] ) % p ; }}int Quick_Pow( int alpha , int beta ){int ans = 1 ; while ( beta > 0 ){if( beta & 1 ) ans = ( ans * alpha ) % mod ; beta >>= 1 ; alpha = ( alpha * alpha ) % mod ; }return ans ; }int Regular_C_of_Pow_Class( int n , int m ){int alpha = 1 , beta = 1 , rereturn = 0 ; if( m <= n && n >= 0 && m >= 0 ){for( int i = n - m + 1 ; i <= n ; ++ i ){alpha = ( alpha * i ) % mod ; }for( int i = 1 ; i <= m ; ++ i ){beta = ( beta * i ) % mod ; } rereturn = ( alpha * Quick_Pow( beta , mod - 2 ) ) % mod ; return rereturn ; }else return 0 ; }inline int jc( int x ){return sum[ x ] ; } inline int neo_jc( int x ){ if( x == 0 ) return 1 ; return sum_neo[ x ] ; }int Regular_C_of_Inv( int n , int m ){return ( ( ( jc( n ) * neo_jc( n - m ) ) % mod ) * neo_jc( m ) ) % mod ; }int C_Lucas_Using_Inv( int n , int m ){if( m == 0 ) return 1 ; return ( Regular_C_of_Inv( n % mod , m % mod ) * C_Lucas_Using_Inv( n / mod , m / mod ) ) % mod ; }int C_Lucas_Using_Pow( int n , int m ){if( m == 0 ) return 1 ; return ( Regular_C_of_Pow_Class( n % mod , m % mod ) * C_Lucas_Using_Pow( n / mod , m / mod ) ) % mod ; }void Asking_for_Derangement( ){D[ 0 ] = 1 ; D[ 1 ] = 0 ; D[ 2 ] = 1 ; for( int i = 3 ; i < N ; ++ i ){D[ i ] = ( i - 1 ) * ( D[ i - 1 ] + D[ i - 2 ] ) % mod ; }}inline void Cleared( ){memset( D , 0 , sizeof( D ) ) ; memset( sum_neo , 0 , sizeof( sum_neo ) ) ; memset( sum , 0 , sizeof( sum ) ) ; memset( nueyong , 0 , sizeof( nueyong ) ) ; }
} ;
using namespace Combination ;
inline int read( )
{int x = 0 , f = 1 ; char c = getchar( ) ; while ( c < '0' || c > '9' ){if( c == '-' ){f = -1 ; }c = getchar( ) ; }while ( c >= '0' && c <= '9' ){x = x * 10 + c - '0' ; c = getchar( ) ; }return x * f ;
}
int l , r , n , t ;
signed main( )
{#ifndef ONLINE_JUDGEfreopen( "cjs.in" , "r" , stdin ) ; freopen( "cjs.out" , "w" , stdout ) ; #endift = read( ) ;while ( t -- ){n = read( ) ; l = read( ) ; r = read( ) ; int m = r - l + 1 ; cout << ( C_Lucas_Using_Pow( n + m , n ) - 1 + mod ) % mod << '\n' ; }
}
