HackerEarth Large Sub-Arrays problem solution YASH PAL, 31 July 2024 In this HackerEarth Large Sub-Arrays problem solution You are given an array A with size N and two integers M and K. Let’s define another array B with size N x M as the array that’s formed by concatenating M copies of array A. You have to find the number of sub-arrays of the array B with sum <= K. Since the answer can be very large you have to print the answer mod 10^9+7. HackerEarth Large Sub-Arrays problem solution. #include <bits/stdc++.h>#include <tr1/unordered_map>using namespace std;using namespace std::tr1;#define opt ios_base::sync_with_stdio(0)#define lli long long int#define ulli unsigned long long int#define I int#define S string#define D double#define rep(i,a,b) for(i=a;i<b;i++)#define repr(i,a,b) for(i=a;i>b;i--)#define in(n) scanf("%lld",&n)#define in2(a,b) scanf("%lld %lld",&a,&b)#define in3(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)#define out(n) printf("%lldn",n)#define inu(a) scanf("%lld",&a)#define outu(a) printf("%llu",a)#define ins(s) scanf("%s",&s)#define outs(s) printf("%s",s)#define mod 1000000007#define inf 100000000000000typedef long long ll;typedef pair<lli,lli> plli;typedef vector<lli> vlli;typedef vector<ulli> vulli;typedef vector<ll> vll;typedef vector<string> vs;typedef vector<plli> vplli;#define MM(a,x) memset(a,x,sizeof(a));#define ALL(x) (x).begin(), (x).end()#define P(x) cerr<<"{"#x<<" = "<<(x)<<"}"<<endl;#define P2(x,y) cerr<<"{"#x" = "<<(x)<<", "#y" = "<<(y)<<"}"<<endl;#define P3(x,y,z) cerr<<"{"#x" = "<<(x)<<", "#y" = "<<(y)<<", "#z" = "<<(z)<<"}"<<endl;#define P4(x,y,z,w)cerr<<"{"#x" = "<<(x)<<", "#y" = "<<(y)<<", "#z" = "<<(z)<<", "#w" = "<<(w)<<"}"<<endl;#define PP(x,i) cerr<<"{"#x"["<<i<<"] = "<<x[i]<<"}"<<endl;#define TM(a,b) cerr<<"{"#a" -> "#b": "<<1000*(b-a)/CLOCKS_PER_SEC<<"ms}n";#define UN(v) sort(ALL(v)), v.resize(unique(ALL(v))-v.begin())#define mp make_pair#define pb push_back#define f first#define s second#define sz() size()#define nl cout<<"n"#define MX1 100005#define MX2 1000005#define bs binary_search#define lb lower_bound#define ub upper_boundlli dx[]={0,0,-1,1,-1,-1,1,1};lli dy[]={1,-1,0,0,1,-1,-1,1};lli power(lli a,lli b) { lli value; if(b==0) { return 1; } else if(b%2==0) { value=power(a,b/2)%mod; return(value*value)%mod; } else { value=power(a,b/2)%mod; return ((a*value)%mod*(value))%mod; }}string f(lli n) { vlli v; lli i; if(n==0) { return "00"; } string s="0"; while(n) { v.pb(n%10); n/=10; } reverse(ALL(v)); rep(i,0,v.sz()) { s+=(v[i]+'0'); } return s;}lli a[100005],pref[100005],suf[100005];int main(){ /*lli files=10; while(files--) { #ifndef ONLINE_JUDGE string input="in"+f(files)+".txt"; string output="out"+f(files)+".txt"; freopen(input.c_str(),"r",stdin); freopen(output.c_str(),"w",stdout); #endif*/ opt; lli t; cin>>t; while(t--) { lli n,m,k; cin>>n>>m>>k; lli ans=0,i,sum=0; rep(i,1,1+n) { cin>>a[i]; pref[i]=a[i]; suf[i]=a[i]; sum+=a[i]; } sum*=m; if(sum<=k) { lli temp=(n*m)%mod; ans=(temp*(temp+1))%mod; ans=(ans*power(2,mod-2))%mod; cout<<ans<<endl; continue; } sum/=m; rep(i,2,1+n) { pref[i]+=pref[i-1]; } repr(i,n-1,0) { suf[i]+=suf[i+1]; } rep(i,1,1+n) { if(suf[i]>=k) { lli temp=-1,l=i,r=n; while(l<=r) { lli mid=(l+r)/2; if((pref[mid]-pref[i-1])<=k) { temp=mid; l=mid+1; } else { r=mid-1; } } if(temp!=-1) { ans=(ans+(m*(temp-i+1)))%mod; } continue; } if((suf[i]+(m-1)*sum)<=k) { lli temp=(m*(2*(n-i+1)+(m-1)*n)); temp/=2; ans=(ans+temp)%mod; continue; } lli cnt=(k-suf[i])/sum; cnt++; lli temp=(cnt*(2*(n-i+1)+(cnt-1)*n)); temp/=2; temp%=mod; ans=(ans+temp)%mod; cnt--; lli val=(k-suf[i]-cnt*sum); lli l=1,r=n,idx=0; while(l<=r) { lli mid=(l+r)/2; if(pref[mid]<=val) { idx=mid; l=mid+1; } else { r=mid-1; } } lli no=(n-i+1)+(cnt*n)+idx; no%=mod; ans=(ans+(no*(m-cnt-1)))%mod; } cout<<ans<<endl; }} coding problems