Sharing training greedy card Noip2002

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Topic links: https://www.luogu.com.cn/problem/P1031

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Title Description

   There N stack of cards, are numbered  1,2, ..., N . There are a number of sheets per stack, but the total number of cards will be for the N factor. A plurality of cards can take on either a pile and then moved.

   Card moving rules: the number of a pile of cards taken, only moved numbered heap 2; numbered heap take N card can only move number N - 1 of the heap;

　　　　　　   Take another heap of cards, can be moved left or right adjacent to the heap.

   Now asked to find a mobile method with the least number of moves are the number of cards per heap as much.

   E.g.  N = . 4, . 4 stack number cards are:

   ①9  ②8  ③17  ④6

   Mobile 3 can achieve:

   Taken from ③ . 4 cards into ④ ( 9,8,13,10 );

   Taken from ③ . 3 cards into ② ( 9,11,10,10 );

   Take from ② 1 card into ① ( 10,10,10,10 );

Input Format

   Two lines

   First line: N ( N card stack, . 1 ≤ N ≤ . 1 0 0)

   Second line: A . 1 , A 2 , ... , A n-  ( N stacks of cards, each card stack the initial number, . 1 ≤ A I ≤ . 1 0 0 0 0)

Output Format

   Line: that is, all heap meet the minimum number of moves equal.

Sample input and output

　Input # 1

. 4 
. 9. 8. 17. 6 

output # 1

3

. 1 #include <bits / STDC ++ H.>
 2  the using  namespace STD;
 . 3  const  int MAXN = 10001 ;
 . 4  int A [MAXN];
 . 5  int main () {
 . 6      int n-, SUM = 0 , Ave = 0 ;
 . 7      CIN> > n-;
 . 8      for ( int I = 0 ; I <n-; I ++ ) {
 . 9          CIN >> a [I];                     // read the data when calculating the average of 
10          Ave + = a [I]; 
 . 11      } 
 12 is     Ave / = n-;
 13 is      for ( int I = 0 ; I <n-; I ++ ) {
 14          A [I] - = Ave;                     // with each subtracting the average, the final result is calculated with the difference between the number 
15      }
 16      for ( int I = 0 ; I <n-; I ++ ) {
 . 17          IF (! A [I] = 0 ) {
 18 is              A [I + . 1 ] = A [I] + A [I + . 1 ];         // Look front most solution, but assuming rightward added, then after moving a [i + 1] is a front plus the original number 
. 19              a [i] == 0 ;                 // if the post-movement position i is 0 above purpose
20             sum++;
21         }else{
22             continue;
23         }
24     } 
25     cout<<sum<<endl;
26     return 0;
27 }