【HDU】4928 Series 2 模拟

本文主要是介绍【HDU】4928 Series 2 模拟，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

传送门：【HDU】4928 Series 2

题目分析：

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 travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )
#define CLR( a , x ) memset ( a , x , sizeof a )typedef long long LL ;const int MAXN = 100005 ;int n ;
LL a[MAXN] ;bool check ( int l , int r ) {bool flag1 = 1 , flag2 = 1 ;FOR ( i , l + 1 , r ) {if ( a[i - 1] > a[i] ) flag1 = 0 ;if ( a[i - 1] < a[i] ) flag2 = 0 ;}return flag1 || flag2 ;
}void scanf ( LL& x , char c = 0 , bool flag = 0 ) {while ( ( c = getchar () ) != '-' && ( c < '0' || c > '9' ) ) ;if ( c == '-' ) flag = 1 , x = 0 ;else x = c - '0' ;while ( ( c = getchar () ) >= '0' && c <= '9' ) x = x * 10 + c - '0' ;if ( flag ) x = -x ;
}void solve () {scanf ( "%d" , &n ) ;int l = 0 , r = n - 1 ;REP ( i , 0 , n ) scanf ( a[i] ) ;REP ( i , 0 , n ) {while ( l < r && !a[l] ) ++ l ;while ( l < r && !a[r] ) -- r ;if ( l > 0 ) -- l ;if ( r < n - i - 1 ) ++ r ;if ( !check ( l , r ) ) {if ( i ) printf ( "%d\n" , i - 1 ) ;else printf ( "ugly series\n" ) ;return ;} else if ( r - l <= 2 ) {printf ( "nice series\n" ) ;return ;}REP ( j , l , r ) a[j] = a[j + 1] - a[j] ;-- r ;}
}int main () {int T ;scanf ( "%d" , &T ) ;while ( T -- ) solve () ;return 0 ;
}

这篇关于【HDU】4928 Series 2 模拟的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！

---