uva11300Spreading the Wealth<数学>

链接:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2275

思路

原来金币数为a1,a2,`````an;

最终的金币为这些数的平均值，设为M

xi表示i给i+1的金币个数

a1+xn-x1=M

a2+x1-x2=M

a3+x2-x3=M

a4+x3-x4=M

.......

an+x(n-1)-xn=M

利用后面n-1个等式，用x1表示 x2,x3,...xn

x2=x1-(M-a2)

x3=x1-(2*M-a2-a3);

....



结果就是求|x1|+|x1-b1|+|x1-b2|+...+|x1-b[n-1]|的最小值，取中位数;

View Code

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <string>

 4 #include <cstring>

 5 #include <cmath>

 6 #include <algorithm>

 7 using namespace std;

 8 typedef long long LL;

 9 int N;

10 LL c[1000005], sum, M, a;

11 LL Labs(  LL a )

12 {

13     return a>0?a:-a;

14 }

15 int main( )

16 {

17     while( scanf("%d", &N )!= EOF ){

18         sum=0, c[0]=0;

19         for( int i=1; i<=N; ++ i ){

20             scanf( "%lld", &a );

21             sum+=a;    

22             c[i]=sum;

23         }

24         M=sum/N;

25         sum=0;

26         for( int i=1; i<=N; ++ i ){

27             c[i]-=i*M;

28         }

29         sort( c, c+N );

30         M=c[N/2];

31         for( int i=0; i<N; ++ i ){

32             c[i]-=M;

33             sum+=Labs(c[i]);

34         }

35         printf( "%lld\n", sum );

36     }

37     return 0;

38 }