Problem Link : http://www.spoj.com/problems/KALTSUM/
Solution Idea:
Break the problem into two cases: When k is greater than sqrt(n) and when k is less tahn or equal sqrt(n). Deal with them in different ways.
Case 1: For any k>sqrt(N) there will be less than sqrt(N) alternating segment. If you have an array, having cumulative sum, you can get the sum of a contiguous segment in O(1). So you can just loop over the alternating parts and get the sum in O(sqrt(N)) for a single query.
Case 2: Now when k <= sqrt(n), you need to do some preprocessing.
let arr[k][i] be the "k alternating sum" of a subarray which starts at position i and keeps alternating untill it reaches a position x such that if you add another segment of size k then it will go out of the array. [ This array can be computed in O(N) , and as you're doing it for every k < sqrt(n), the total complexity is O(N * sqrt(N))] With the help of this array, you can answer every query having k < sqrt(N) in O(1).
This solution idea is from this link.
#include <bits/stdc++.h>
#define pii pair <int,int>
#define pll pair <long long,long long>
#define sc scanf
#define pf printf
#define Pi 2*acos(0.0)
#define ms(a,b) memset(a, b, sizeof(a))
#define pb(a) push_back(a)
#define MP make_pair
#define db double
#define ll long long
#define EPS 10E-10
#define ff first
#define ss second
#define sqr(x) (x)*(x)
#define D(x) cout<<#x " = "<<(x)<<endl
#define VI vector <int>
#define DBG pf("Hi\n")
#define MOD 1000000007
#define CIN ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a) (int)a.size()
#define sf(a) scanf("%d",&a)
#define sfl(a) scanf("%lld",&a)
#define sff(a,b) scanf("%d %d",&a,&b)
#define sffl(a,b) scanf("%lld %lld",&a,&b)
#define sfff(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v) for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n) for(int i=0;i<n;i++)
#define loop1(i,n) for(int i=1;i<=n;i++)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define RREP(i,a,b) for(int i=a;i>=b;i--)
#define TEST_CASE(t) for(int z=1;z<=t;z++)
#define PRINT_CASE printf("Case %d: ",z)
#define LINE_PRINT_CASE printf("Case %d:\n",z)
#define CASE_PRINT cout<<"Case "<<z<<": "
#define all(a) a.begin(),a.end()
#define intlim 2147483648
#define infinity (1<<28)
#define ull unsigned long long
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a)*((b)/gcd(a,b)))
using namespace std;
/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1}; // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1}; // Kings Move
//const int fx[]={-2, -2, -1, -1, 1, 1, 2, 2}; // Knights Move
//const int fy[]={-1, 1, -2, 2, -2, 2, -1, 1}; // Knights Move
/*------------------------------------------------*/
/*-----------------------Bitmask------------------*/
//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));}
/*------------------------------------------------*/
ll dp[318][100005];
ll csum[100005];
ll ara[100005];
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int n,q;
sff(n,q);
for(int i=1;i<=n;i++)
{
sfl(ara[i]);
csum[i]=csum[i-1]+ara[i];
}
int root=ceil(sqrt(n));
for(int i=1;i<=root;i++)
{
ll temp=0,sub=0;
for(int j=n;j>=1;j--)
{
temp+=ara[j];
if(j+i>n)
sub=0;
else
sub=ara[j+i];
temp-=sub;
if(j+i-1>n)
dp[i][j]=0;
else
dp[i][j]=temp-dp[i][j+i];
}
}
//
// for(int i=1;i<=root;i++,cout<<endl)
// for(int j=1;j<=n;j++)
// cout<<dp[i][j]<<" ";
while(q--)
{
int a,b,k;
sfff(a,b,k);
ll ans=0;
if(k>root)
{
for(int i=a,j=0;i<=b;i+=k,j++)
{
if(j%2==0)
ans+=csum[i+k-1]-csum[i-1];
else
ans-=csum[i+k-1]-csum[i-1];
}
pf("%lld\n",ans);
}
else
{
int turn=(b-a+1)/k;
if(turn%2==1)
pf("%lld\n",dp[k][a]+dp[k][b+1]);
else
pf("%lld\n",dp[k][a]-dp[k][b+1]);
}
}
return 0;
}