https://codeforces.com/problemset/problem/792/E
最后肯定是全是x,x+1
对于颜色i,如果它分成的set数>=sqrt(ai),是一定可行的,因为ai%sqrt(a[i])<sqrt(a[i])<=ai/sqrt(a[i])
所以我们枚举最小的ai分成的set数k 1到sqrt(a[i]),这样就可以确定x,然后去check(x),如果ai%k==0,也要check(x-1)
#include <bits/stdc++.h> #define inf 2333333333333333 #define N 1000010 #define p(a) putchar(a) #define For(i,a,b) for(long long i=a;i<=b;++i) //by war //2020.11.20 using namespace std; long long n,t,x,ans; long long a[N]; void in(long long &x){ long long y=1;char c=getchar();x=0; while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();} while(c<='9'&&c>='0'){ x=(x<<1)+(x<<3)+c-'0';c=getchar();} x*=y; } void o(long long x){ if(x<0){p('-');x=-x;} if(x>9)o(x/10); p(x%10+'0'); } long long check(long long x){ long long tot=0; For(i,1,n){ if(a[i]/x>=a[i]%x) tot+=a[i]/x-(a[i]/x-a[i]%x)/(x+1); else return inf; } return tot; } signed main(){ in(n); For(i,1,n) in(a[i]); sort(a+1,a+n+1); t=sqrt(a[1]);ans=inf; For(i,1,t){ x=a[1]/i; if(a[1]%i==0 && x>1) ans=min(ans,check(x-1)); ans=min(ans,check(x)); } o(ans); return 0; }