Problem:
Score : 500 points

Problem Statement

Given are a sequence of N positive integers A_1, A_2, \ldots, A_N, and a positive integer K.

Find the number of non-empty contiguous subsequences in A such that the remainder when dividing the sum of its elements by K is equal to the number of its elements. We consider two subsequences different if they are taken from different positions, even if they are equal sequences.

Constraints

All values in input are integers.

1 \leq N \leq 2\times 10^5

1 \leq K \leq 10^9

1 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 \cdots A_N

Output

Print the number of subsequences that satisfy the condition.

Sample Input 1

5 4
1 4 2 3 5

Sample Output 1

4

Four sequences satisfy the condition: (1), (4,2), (1,4,2), and (5).

Sample Input 2

8 4
4 2 4 2 4 2 4 2

Sample Output 2

7

(4,2) is counted four times, and (2,4) is counted three times.

Sample Input 3

10 7
14 15 92 65 35 89 79 32 38 46

Sample Output 3

8

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>

using namespace std;

#define MAX             100005
#define MOD             1000003
#define eps             1e-6
int fx[] =              {1,-1,0,0};
int fy[] =              {0,0,1,-1};

#define FastRead        ios_base::sync_with_stdio(0);cin.tie(0);
#define fRead           freopen("in.txt","r",stdin);
#define fWrite          freopen ("out.txt","w",stdout);

#define ll              long long
#define ull             unsigned long long
#define ff              first
#define ss              second
#define pb              push_back
#define PI              acos(-1.0)
#define mk              make_pair
#define pii             pair<int,int>
#define pll             pair<LL,LL>
#define vi              vector <int>
#define vl              vector <LL>
#define vp              vector<pii>
#define all(a)          a.begin(),a.end()

#define min3(a,b,c)     min(a,min(b,c))
#define max3(a,b,c)     max(a,max(b,c))
#define min4(a,b,c,d)   min(a,min(b,min(c,d)))
#define max4(a,b,c,d)   max(a,max(b,max(c,d)))

#define FOR(i,a,b)      for(int i=a;i<=b;i++)
#define ROF(i,a,b)      for(int i=a;i>=b;i--)
#define REP(i,b)        for(int i=0;i<b;i++)
#define IT(it,x)	for(it=x.begin();it!=x.end;it++)

#define MEM(a,x)        memset(a,x,sizeof(a))
#define ABS(x)          ((x)<0?-(x):(x))
#define SORT(v)         sort(v.begin(),v.end())
#define REV(v)          reverse(v.begin(),v.end())

//#define NUM 1000005bool flag[NUM];vector <int> prime;inline void sieve(){flag[1]=true;for(int i=4; i<NUM; i+=2){flag[i]=true;}for(int i=3; i*i<=NUM; i+=2){if(!flag[i]){for(int j=i*i; j<NUM; j+=2*i){flag[j]=true;}}}prime.pb(2);for(int i=3; i<NUM; i+=2){if(!flag[i]){prime.pb(i);}}return;}

//inline int NOD(ll n){int div=1;for(int i=0; prime[i]*prime[i]<=n; i++){if(n%prime[i]==0){int cnt=0;while(n%prime[i]==0){n/=prime[i];cnt++;}div*=(cnt+1);}}if(n>1){div*=2;}return div;}

//inline ll BigMod(ll B,ll P,ll M){ll R=1;while(P>0){if(P & 1) R=(R*B)%M;P/=2;B=(B*B)%M;}return R%M;}

//inline void build(int L,int R,int pos){if(L==R){tree[pos]=arr[L];return;}int mid=(L+R)/2;build(L,mid,pos*2+1);build(mid+1,R,pos*2+2);tree[pos]=min(tree[pos*2+1],tree[pos*2+2]);return;}

//inline int query(int ql,int qr,int L,int R,int pos){if(ql>R or qr<L) return MAX;else if(ql<=L and qr>=R) return tree[pos];int mid=(L+R)/2;int p=query(ql,qr,L,mid,2*pos+1);int q=query(ql,qr,mid+1,R,2*pos+2);return min(p,q);}

//void updateRange(int b,int e,int L,int R,int pos,ll val){if(lazy[pos]!=0){tree[pos]+=(R-L+1)*lazy[pos];if(L!=R){lazy[2*pos+1]+=lazy[pos];lazy[2*pos+2]+=lazy[pos];}lazy[pos]=0;}if(L>R or L>e or R<b) return;if(L>=b and R<=e){tree[pos]+=(R-L+1)*val;if(L!=R){lazy[2*pos+1]+=val;lazy[2*pos+2]+=val;}return;}int mid=(L+R)/2;updateRange(b,e,L,mid,2*pos+1,val);updateRange(b,e,mid+1,R,2*pos+2,val);tree[pos]=tree[2*pos+1]+tree[2*pos+2];return;}

//ll getSum(int ql,int qr,int L,int R,int pos){if(lazy[pos]!=0){tree[pos]+=(R-L+1)*lazy[pos];if(L!=R){lazy[2*pos+1]+=lazy[pos];lazy[2*pos+2]+=lazy[pos];}lazy[pos]=0;}if(L>R or ql>R or qr<L) return 0;if(L>=ql and qr>=R) return tree[pos];int mid=(L+R)/2;return getSum(ql,qr,L,mid,2*pos+1)+getSum(ql,qr,mid+1,R,2*pos+2);}

//inline void bfs(int s){MEM(vis,0);vis[s]=1;queue<int>Q;Q.push(s);while(!Q.empty()){int x=Q.front();Q.pop();REP(i,V[x].size()){if(!vis[V[x][i]]){vis[V[x][i]]=1;Q.push(V[x][i]);}}}return;}

//int cell[100][100]; int d[100][100],vis[100][100];int row,col;inline void bfs(int sx,int sy){ MEM(vis,0);vis[sx][sy]=1;queue<pii>q; q.push(pii(sx,sy));while(!q.empty()){pii top=q.front();q.pop();for(int k=0; k<4; k++){int tx=top.ff+fx[k];int ty=top.ss+fy[k]; if(tx>=0 and tx<row and ty>=0 and ty<col and cell[tx][ty]!=-1 and vis[tx][ty]==0){vis[tx][ty]=1;d[tx][ty]=d[top.ff][top.ss]+1;q.push(pii(tx,ty)); }}}return;}

//inline void bi_dfs(int s,int c){if(vis[s]) return;vis[s]=1;color[s]=c;REP(i,V[s].size()) dfs(V[s][i],1-c);return;}

//struct point{int name,val;bool operator <(const point &p) const{return p.val < val;}};vector<int>V[105];int dis[105],cost[105][105];priority_queue<point>Q;void Dijkstra(int s){dis[s]=0;point get;get.name=s;get.val=0;Q.push(get);while(!Q.empty()){point tmp=Q.top();Q.pop();int now=tmp.name;REP(i,V[now].size()){int x=V[now][i];if(dis[now]+cost[now][x]<dis[x]){dis[x]=dis[now]+cost[now][x];get.name=x;get.val=dis[x];Q.push(get);}}}return;}

//struct edge{int u,v,w;bool operator < (const edge &p) const{return w < p.w;}};int pr[MAX];vector<edge>e;int find(int r){return pr[r]= (pr[r]==r) ? r:find(pr[r]);}int kruskal(int n){sort(e.begin(),e.end());FOR(i,1,n) pr[i]=i;int cnt=0,sum=0;REP(i,e.size()){int x=find(e[i].u);int y=find(e[i].v);if(x!=y){pr[x]=y;cnt++;sum+=e[i].w;if(cnt==n-1) break;}}return sum;}

//inline ll GCD (ll a,ll b ) {if(a < b) swap(a, b); while ( b ) { a = a % b; swap ( a, b ); } return a;}

//int Set(int N,int pos){return N=N | (1<<pos);}

//int reset(int N,int pos){return N= N & ~(1<<pos);}

//bool check(int N,int pos){return (bool)(N & (1<<pos));}

//// ***************************************************** TEMPLATE ***************************************************** ////

ll sum[200005];
ll arr[200005];

map <ll,ll> rem;

int main()
{
    ll n,k;
    cin >> n >> k;

    ll ans=0;

    rem[0]=1;

    FOR(i,1,n)
    {
        cin >> arr[i];
        sum[i]=(sum[i-1]+arr[i])%k;
        if(i-k>=0)
        {
            rem[(sum[i-k]-(i-k))%k]--;
        }
        ans+=rem[(sum[i]-i)%k];
        //cout << i << ' ' << rem[(sum[i]-i)%k] << '\n';
        rem[(sum[i]-i)%k]++;
    }

    cout << ans;

}