# HackerEarth Partitions problem solution

In this HackerEarth Partitions problem solution, you are given an array a of size n and two integers l and r. You have to find the number of partitions of array a such that the sum of elements in each partition lies between  and (both inclusive). Since the answer can be large, find the answer modulo 10^9 + 7 (1000000007).

## HackerEarth Partitions 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("%lld\n",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;    }}lli a[100005],dp[100005],n,L,R,dp_brute[100005],tree[400005],prefa[100004],check=0;void update(lli n,lli s,lli e,lli idx,lli val) {    if(s==e) {        tree[n]=val;    } else {        lli mid=(s+e)/2;        if(idx<=mid) {            update(2*n,s,mid,idx,val);        } else {            update(2*n+1,mid+1,e,idx,val);        }        tree[n]=(tree[2*n]+tree[2*n+1])%mod;;    }}lli query(lli n,lli s,lli e,lli l,lli r) {    if(s>e or s>r or e<l) {        return 0;    }    if(s>=l and e<=r) {        return tree[n]%mod;    }    lli mid=(s+e)/2;    return (query(2*n,s,mid,l,r)+query(2*n+1,mid+1,e,l,r))%mod;}int main(){    opt;    cin>>n>>L>>R;    lli i,j,flag=0;    rep(i,1,1+n) {        cin>>a[i];        if(a[i]>R) {            flag=1;        }        prefa[i]=prefa[i-1]+a[i];    }    if(flag) {        cout<<0<<endl;        exit(0);    }    update(1,1,n+1,n+1,1);    lli sumL=0,sumR=0,add=1,l=n-1,r=n-1;    repr(i,n,0) {        lli st=-1,en=-1;        lli l=i,r=n;        while(l<=r) {            lli mid=(l+r)/2;            if((prefa[mid]-prefa[i-1])>=L) {                st=mid;                r=mid-1;            } else {                l=mid+1;            }        }        l=i,r=n;        while(l<=r) {            lli mid=(l+r)/2;            if((prefa[mid]-prefa[i-1])<=R) {                en=mid;                l=mid+1;            } else {                r=mid-1;            }        }        if(st!=-1 and en!=-1) {            update(1,1,n+1,i,query(1,1,n+1,st+1,en+1)%mod);        }    }    cout<<dp[1]<<endl;}`

### Second solution

`#include<bits/stdc++.h>#define int long longusing namespace std;const int maxn = 1e5 + 100;const int mod = 1e9 + 7;int l, r, n, ar[maxn], dp[maxn], sum_a[maxn], sum_dp[maxn];vector<int> vc;int32_t main(){    cin >> n >> l >> r;    for(int i = 0; i < n; i++)    cin >> ar[i];    for(int i = 0; i < n; i++){    sum_a[i + 1] = sum_a[i] + ar[i];        int st = lower_bound(vc.begin(), vc.end(), sum_a[i + 1] - r) - vc.begin();    int en = upper_bound(vc.begin(), vc.end(), sum_a[i + 1] - l) - vc.begin();    dp[i] = sum_dp[en] - sum_dp[st];    if(sum_a[i+1] >= l && sum_a[i+1] <= r)        dp[i]++;        dp[i] %= mod;      sum_dp[i + 1] = sum_dp[i] + dp[i];    vc.push_back(sum_a[i + 1]);    }    cout << dp[n - 1] << endl;}`