ZROI#985-白红宇

ZROI#985

阅读量：5289 次

发布时间：2019-06-14

本文共 2409 字，大约阅读时间需要 8 分钟。

暴力就不说了,说说正解吧.

先假定每个区间都没有重复元素,然后得到一个全集的答案.

然后我们考虑,减掉不合法的方案.

记录每种颜色出现的位置,乘法原理即可.

暴力\(Code:\)

#include 
    
     #include 
     
      #include 
      
       #define rint read
       
        #define int long longtemplate < class T >    inline T read () {        T x = 0 , f = 1 ; char ch = getchar () ;        while ( ch < '0' || ch > '9' ) {            if ( ch == '-' ) f = - 1 ;            ch = getchar () ;        }        while ( ch >= '0' && ch <= '9' ) {            x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ;            ch = getchar () ;        }        return f * x ;    }const int N = 1e6 + 100 ;int n , v[N] , ans ;bool mk[3005] ;signed main () {    n = rint () ; for (int i = 1 ; i <= n ; ++ i) v[i] = rint () ;    for (int i = 1 ; i <= n ; ++ i)        for (int j = 1 ; j <= i ; ++ j) {            for (int k = 1 ; k <= n ; ++ k) mk[k] = false ;            for (int k = j ; k <= i ; ++ k)                if ( ! mk[v[k]] ) ++ ans , mk[v[k]] = true ;        }    printf ("%lld\n" , ans ) ;    return 0 ;}

此正解思路与上述思路略有不同.

此代码思路是直接统计.

正解\(Code:\)

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              #include 
               
                #define MEM(x,y) memset ( x , y , sizeof ( x ) )#define rep(i,a,b) for (int i = (a) ; i <= (b) ; ++ i)#define per(i,a,b) for (int i = (a) ; i >= (b) ; -- i)#define pii pair < int , int >#define one first#define two second#define rint read
                
                 #define int long long#define pb push_backusing std::queue ;using std::set ;using std::pair ;using std::max ;using std::min ;using std::priority_queue ;using std::vector ;using std::swap ;using std::sort ;using std::unique ;using std::greater ;template < class T > inline T read () { T x = 0 , f = 1 ; char ch = getchar () ; while ( ch < '0' || ch > '9' ) { if ( ch == '-' ) f = - 1 ; ch = getchar () ; } while ( ch >= '0' && ch <= '9' ) { x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ; ch = getchar () ; } return f * x ;}const int N = 1e6 + 100 ;int pre[N] , last[N] , n , v[N] , ans ;signed main (int argc , char * argv[]) { n = rint () ; rep ( i , 1 , n ) v[i] = rint () ; rep ( i , 1 , n ) { pre[i] = last[v[i]] ; last[v[i]] = i ; } rep ( i , 1 , n ) ans += ( i - pre[i] ) * ( n - i + 1 ) ; printf ("%lld\n" , ans ) ; return 0 ;}