rq457 连续串之和

rq457 连续串之和

#include < iostream >
using   namespace  std;

int  n,s;
int  d,S;
int  c[ 10001 ];
int  main()
{
    scanf( " %d%d " , & n, & s);
    S = (n - 1 ) * n / 2 ;
    d = S - s;
     if (s > S || s <- S || d % 2 == 1 ){
        printf( " NIE " );
         return   0 ;
    }
     for ( int  i = 2 ;i <= n; ++ i)
        c[i] = 1 ;
     int  k;
     for ( int  i = 2 ;i <= n && d; ++ i){
        k = (n - i + 1 ) * 2 ;
         if (d >= k){
            c[i] =- 1 ;
            d -= k;
        }
    }
    k = 0 ;
    printf( " %d\n " ,k);
     for ( int  i = 2 ;i <= n; ++ i){
        k += c[i];
        printf( " %d\n " ,k);
    }
     return   0 ;
}