题目大意:将n个数分解成若干组,如4 = 2+2, 7 = 2+2+3,保证所有组中数字之差<=1。

首先我们能想到找一个最小值x,然后从x+1到1枚举并check,找到了就输出。这是40分做法。

能不能优化?我们发现,若k合法,那么x%k==0或x%(k+1)==0或x%(k-1)==0。

所以枚举倍数就行了,利用贪心找到了就输出。

代码:

#include<cstdio>
#include<algorithm>
using namespace std;
#define ll long long
int n;
ll a[],mn=0x3f3f3f3f;
bool cmp(ll x,ll y)
{
return x<y;
}
void check(ll k)
{
if(!k)return ;
ll ret = ;
for(int i=;i<=n;i++)
{
if(a[i]<k-)return ;
int cnt = a[i]/k+(a[i]%k!=);
if(a[i]%k==)
{
ret+=cnt;
continue;
}
if(k*cnt-a[i]>cnt)return ;
ret+=cnt;
}
printf("%I64d\n",ret);
exit();
return ;
}
int main()
{
// freopen("C.in","r",stdin);
// freopen("C.out","w",stdout);
scanf("%d",&n);
for(int i=;i<=n;i++)
{
scanf("%I64d",&a[i]);
mn=min(mn,a[i]);
}
for(int i=;i<=mn;i++)
{
check(mn/i+);
check(mn/i);
check(mn/i-);
}
fclose(stdin);
fclose(stdout);
return ;
}

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