题意简介明了，需要找到一个(T)，最小化
[sum_{i=1}^nleft lfloor frac{a_i}{T}
ight
floor+sum_{i=1}^na_i\%T
]
非常显然的(a_i\%T=a_i-left lfloor frac{a_i}{T} ight floor imes T)
于是
[sum_{i=1}^nleft lfloor frac{a_i}{T}
ight
floor+sum_{i=1}^na_i-T imes sum_{i=1}^nleft lfloor frac{a_i}{T}
ight
floor
]
即为
[sum_{i=1}^na_i-(T-1)sum_{i=1}^nleft lfloor frac{a_i}{T}
ight
floor
]
最小化这个柿子只需要最大化((T-1)sum_{i=1}^nleft lfloor frac{a_i}{T} ight floor)
考虑一次枚举(T)，需要快速求出(sum_{i=1}^nleft lfloor frac{a_i}{T} ight floor)
注意到(left lfloor frac{a_i}{T} ight floor)只会有(left lfloor frac{max a_i}{T} ight floor)种值，即对于(a_iin[0,T-1],left lfloor frac{a_i}{T} ight floor=0...a_iin [kT-T,kT-1],left lfloor frac{a_i}{T} ight floor=k)
我们直接暴力这(left lfloor frac{max a_i}{T} ight floor)段区间，前缀和算一下这段区间里有多少个(a_i)即可
复杂度显然调和级数，视(n)与(max a_i)同级，复杂度为(O(nlog n))
代码
#include<bits/stdc++.h>
#define re register
#define LL long long
#define max(a,b) ((a)>(b)?(a):(b))
const int maxn=1e6+5;
inline int read() {
char c=getchar();int x=0;while(c<'0'||c>'9') c=getchar();
while(c>='0'&&c<='9') x=(x<<3)+(x<<1)+c-48,c=getchar();return x;
}
int n,pre[maxn],T;LL ans,tmp;
inline int calc(int l,int r) {
return (r>T?pre[T]:pre[r])-(l?pre[l-1]:0);
}
int main() {
n=read();
for(re int x,i=1;i<=n;i++) x=read(),ans+=x,T=max(T,x),pre[x]++;
for(re int i=1;i<=T;i++) pre[i]+=pre[i-1];
for(re int i=2;i<=T;++i) {
LL now=0;
for(re int cnt=0,l=0,j=i-1;l<=T;j+=i,l+=i,++cnt)
now+=1ll*calc(l,j)*cnt;
if(1ll*now*(i-1)>tmp) tmp=1ll*now*(i-1);
}
printf("%lld
",ans-tmp);
return 0;
}