BZOJ 1045 糖果传递(思维)

设第i个人给了第i+1个人mi个糖果(可以为负)，因为最后每个人的糖果都会变成sum/n。

可以得到方程组 mi-mi+1=a[i+1]-sum/n.(1<=i<=n).

把方程组化为mn组成的形式，最后的结果就是求min(abs(mn)+abs(mn-a[i+1]+sum/n)....)。可以看出这是一个分段函数。

且函数最值在mn取中位数的地方。

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-3
# define MOD 1000000007
# define INF (LL)1<<60
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=1000005;
//Code begin...
 
int a[N], b[N];
LL sum[N];
 
int main()
{
    int n;
    LL ave=0, ans=0;
    scanf("%d",&n);
    FOR(i,1,n) scanf("%d",a+i), ave+=a[i];
    ave/=n;
    FOR(i,1,n) a[i]-=ave;
    FOR(i,1,n) sum[i]=a[i]+sum[i-1];
    sort(sum+1,sum+n+1);
    LL t=sum[(1+n)>>1];
    FOR(i,1,n) ans+=abs(t-sum[i]);
    printf("%lld\n",ans);
    return 0;
}