Codeforces Round # 601 (Div. 2) E (looking prime factors, DP)

First decomposition of the quality factor for the current a [i], the quality factor is assumed that the current x, this location to meet the divisible k, there are two possibilities, either it is a rearwardly x% k th transfer, or after a transfer it to a kx% k.

For each of the a [i] is satisfied, because only affect the next position, so that the next position of a [i + 1] Effect count a [i] is generated, and later issues a new sub (prohibition nesting dolls).

A quality factor for each of x, for all i positions all requirements min (a [i], xa [i]) and, the minimum value of the sum of all prime factors of i as the answer.

 1 #define HAVE_STRUCT_TIMESPEC
 2 #include<bits/stdc++.h>
 3 using namespace std;
 4 long long a[1000007];
 5 int main(){
 6     ios::sync_with_stdio(false);
 7     cin.tie(NULL);
 8     cout.tie(NULL);
 9     long long n;
10     cin>>n;
11     long long sum=0;
12     for(long long i=1;i<=n;++i){
13         cin>>a[i];
14         sum+=a[i];
15     }
16     if(sum<=1){
17         cout<<-1;
18         return 0;
19     }
20     long long ans=1e18;
21     for(long long i=2;i*i<=sum;++i){
22         long long flag=0;
23         while(sum%i==0){
24             sum/=i;
25             flag=1;
26         }
27         if(flag){
28             long long pre=0,summ=0;
29             for(long long j=1;j<=n;++j){
30                 pre=(a[j]+pre)%i;
31                 summ+=min(pre,i-pre);
32             }
33             ans=min(ans,summ);
34         }
35     }
 36      if (sum> 1 ) {
 37          along  long pre = 0 , summ = 0 ;
38          for ( long  along j = 1 ; j <= n; ++ j) {
 39              pre = (a [j] + pre)% sum;
40              summ = min + (pre, sum- pre);
41          }
 42          years = min (years summ);
43      }
 44      cout << years;
45      return  0 ;
46 }