Ultraviolet A 11300 spreading the wealth

F. Spreading the wealth

Problem

A communist regime is trying to redistribute wealth in a village. they have decided to sit everyone around und a circular table. first, everyone has converted all of their properties to coins of equal value, such that the total number of coins is divisible by the number of people in the village. finally, each person gives a number of coins to the person on his right and a number coins to the person on his left, such that in the end, everyone has the same number of coins. given the number of coins of each person, compute the minimum number of coins that must be transferred using this method so that everyone has the same number of coins.

The input

There is a number of inputs. Each input beginsN(N<1000001), the number of people in the village.NLines follow, giving the number of coins of each person in the village, in counterclockwise order around the table. The total number of coins will fit inside an unsigned 64-bit integer.

The output

For each input, output the minimum number of coins that must be transferred on a single line.

Sample Input

 
310010010041254
 

Sample output

 
04
 

 

Problem setter: Josh Bao

 

Link: http://uva.onlinejudge.org/external/113/11300.html

Analysis: Good questions. It mainly refers to the analysis process. For details, seeCodeAnnotations.

 /*  The original number of gold coins in the UVA 11300 is A1, A2, '''. the final gold coins are the average values of these numbers, set to Axi to indicate I to I + 1 the number of coins A1 + xn-x1 = Aa2 + x1-x2 = aa3 + x2-x3 = aa4 + x3-x4 = ....... an + x (n-1)-xn = A uses N-1 1 equations and X1 to represent X2, X3 ,... xnx2 = x1-(A-a2) X3 = x1-(2 * A-a2-a3 );.... the result is | X1 | + | x1-b1 | + | x1-b2 | +... + | minimum value of x1-b [n-1] |, median  */  # Include <Stdio. h> # Include <Algorithm> # Include <Iostream> # Include < String . H> # Include <Math. h>Using   Namespace  STD;  Const   Int Maxn = 1000010  ;  Long   Long  A [maxn], B [maxn];  Int  Main (){  Int  N;  While (Scanf ( " % D  " , & N) = 1  ){  Long   Long Sum = 0  ;  For ( Int I = 1 ; I <= N; I ++ ) {Scanf (  "  % I64d  " , & A [I]);//  Use LLD Sum + = A [I];}  Long   Long A = sum/ N; B [  0 ] = 0  ;  For ( Int I = 1 ; I <n; I ++ ) B [I] = B [I- 1 ] + A-A [I +1  ]; Sort (B, B + N );  Long   Long T = B [N/ 2  ];  Long   Long Ans = 0  ;  For ( Int I = 0 ; I <n; I ++) ans + = ABS (t- B [I]); printf ( "  % I64d \ n  " , ANS ); //  Use LLD  }  Return   0  ;}