Progression
Title Description
Give you a length \ (N \) in the sequence of positive integers, if a continuous sequence, promoter sequence and can be \ (K \) is divisible, then depending on this sub-sequence legitimate, find the original sequence, including how many legitimate the continuous sequence? For a sequence of length 8, \ (K =. 4 \) case: 2, 1, 2, 1, 1, 2, 1, 2. It answers to 6, sequence is position 1-> position 8, a 2> 4,2-> 7,3-> 5,4> 6,5> 7.
Input formats:
First line: \ (T \) , represents the number of data sets
for each group of data:
The first line: the number 2, \ (K \) , \ (N \)
the second row: \ (N \) number, this sequence represents the
Output formats:
Total T lines, each line represents a number of answers
Sample input:
2 7 3 1 2 3 4 8 2 1 2 1 1 2 1 2
Sample output:
0 6
data range:
100% data satisfies
\ (. 1 <= T <= 20 is \)
\ (. 1 <= N <= 50000 \)
\ (. 1 <= K <= 1000000 \)
\ (each number sequence <= 1000000000 \)
30 % data satisfies
\ (. 1 <= T <= 10 \)
\ (. 1 <= N, K <= 1000 \)
time limit:
1S
Space limitations:
256M
prompt:
remove!!!
answer
And be contiguous subsequence \ (K \) number divisible?
\ (n ^ 2 \) violence? \ (. 1 <= N <= 50000 \) , seemingly not pass.
If we look at some of the maintenance properties of some numbers before this bunch of numbers, each adding a number to update some of these properties. (Do not ask how I think, would have thought dp or greedy, then I kept thinking thought)
every time you add a number, just find this number in front of some of this points to a consecutive sequence of numbers can be \ ( K \) number divisible.
So all we need is the value before the current number and suffix.
We can use an array \ (kk [i] \) represents the die and the suffix \ (K \) in the sense of extension and \ (I \) number, a number is added every time when a determined number \ ( X \) the number of added plus \ (X \) post-mold \ (K \) is 0, so that \ (ANS \) plus \ (kk [x] \) on the line.
But we found update to add a new number \ (kk \) time required is \ (O (n) \) , and
here we use a \ (SUM \) , every time you add a new number\ (SUM \) to add this number, then the \ (kk [i] \) on the die and suffixes \ (K \) in the sense of extension and \ (i + sum% K \ ) number.
Then to add a new number \ (kk \) becomes when \ (kk [(a-sum )% K] ++ \) a. ( \ (A \) as the new number)
to do the time complexity seems to be \ (O (n) \) is.
On the code:
#include<bits/stdc++.h>
using namespace std;
int t,k,n,a[50009];
int sum;
int ans;
int kk[1000009];
int main(){
scanf("%d",&t);
while(t--){
memset(kk,0,sizeof(kk));
ans=sum=0;
scanf("%d%d",&k,&n);
for(int j=1;j<=n;j++){
scanf("%d",&a[j]);
a[j]%=k;
ans+=kk[(k-((a[j]+sum)%k))%k];//(a[j]+sum)%k可能是0，所以k-(a[j]+sum)%k可能为k，所以最后还要%k
sum+=a[j];
sum%=k;
kk[(a[j]-sum+k)%k]++;
if(a[j]%k==0) ans++;//要注意序列里只有一个点的时候也要加到$ans$里去
}
printf("%d\n",ans);
}
return 0;
}