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Birthday Gift Again CodeChef Solution in C++17

``````#include <bits/stdc++.h>
#define ll long long int
#define mod 1000000007
using namespace std;

int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL); cout.tie(NULL);
ll t; cin>>t;
while(t--) {
string s; cin>>s;
ll n = s.size();
vector<ll>one(n+1), zero(n+1);

one[n]=0; zero[n]=0;
for(ll i=n-1; i>=0; i--) {
zero[i] = (s[i] == '0') + zero[i+1];
one[i] = (s[i] == '1') + one[i+1];
}

// cout<<"0: "; for(ll i=0; i<zero.size(); i++) cout<<zero[i]<<" ";
// cout<<"\n1: "; for(ll i=0; i<zero.size(); i++) cout<<one[i]<<" ";
// cout<<endl;

ll idx = 1, ans=0;
while(idx*(idx+1) <= n) {
for(ll i=0; i<=(n-(idx*idx + idx)); i++) {
ll count_zero = zero[i] - zero[i+idx*idx + idx];
ll count_one = one[i] - one[i+idx*idx + idx];

// cout<<i<<" "<<i+idx*idx + idx<<endl;

ans += (count_zero == count_one*count_one);
} idx++;
} cout<<ans<<endl;

}
}``````

Birthday Gift Again CodeChef Solution in C++14

``````#include<bits/stdc++.h>
using namespace std;
#define fastIO ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
typedef long long int lli;
typedef long long int ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef vector<pll> vop;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<int> vi;
typedef pair<int, int> pii;
#define REP(i,a,b) for( long long i=a;i<b;i++)
#define REPR(i,a,b) for(long long i=a;i>=b;i--)
#define adb(a) for(auto ii:a) cout << ii << " "; cout << endl
#define adp(a) for(auto ii:a) cout << ii.F << " " << ii.S << endl;
#define PB push_back
#define MP make_pair
#define F first
#define S second
#define all(x) (x).begin(),(x).end()
#define endl "\n"
#define mem(a,b) memset(a,b,sizeof(a))
#define mod 1000000007
#define PI 3.14159265358979323846
const ll MAXN = 1000001;
//const ll p = 1000000007;
#define N 200005
#define INF INTMAX_MAX

struct hash_pair {
template <class T1, class T2>
size_t operator()(const pair<T1, T2>& p) const
{
auto hash1 = hash<T1> {}(p.first);
auto hash2 = hash<T2> {}(p.second);
return hash1 ^ hash2;
}
};
struct cmp
{
bool operator()(const pll &a, const pll &b)
{
if (a.F == b.F) return a.S > b.S;
return a.F > b.F;
}
};

ll dx[] = {1, -1, 0, 0};
ll dy[] = {0, 0, 1, -1};

ll power(ll a, ll b, ll m) { ll ans = 1; a = a % m; while (b) { if (b & 1) ans = (ans * a) % m; b /= 2; a = (a * a) % m; } return ans; }

ll lcm(ll a, ll b) { return (a * b) / __gcd(a, b); }

ll modInverse(ll a, ll m) { ll g = __gcd(a, m); return power(a, m - 2, m); }

int a[20];
int main()
{
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output2.txt", "w", stdout);
#endif

fastIO;
int test = 1, num = 1;
cin >> test;
while (test--)
{
string s;
cin >> s;
int n = s.size();

int cnt = 0;
int ans = 0;
//map<int,int> mapi;
int j;
vector<int>a(n, 0);
if (s[0] == '1')
a[0]++;
for (int i = 1; i < n; i++) {

a[i] = a[i - 1] + (s[i] == '1');
}
// for (int i = 0; i < n; i++)
//   cout << a[i];
//int ans=0;
for (int i = 0; i < n; i++) {
//previ=a[i];
for (j = 1; j * j + j + i <= n; j++) {
int g;
if (i == 0)
g = 0;
else
g = a[i - 1];
if (a[j * j + j + i - 1] - g == j) {
ans++;
}
}
//   cout << ans;
}
cout << ans << endl;

}
return 0;
}``````

Birthday Gift Again CodeChef Solution in PYTH 3

``````a=int(input())
for i in range(a):
S=input()
N=len(S)
B=[]
x=0
for j in S:
if j=="1":
x+=1
B.append(x)
ans=0
k=1
while (k**2)+k<=N:
p=(k**2)+k-1
x=B[p]
j=p
t=0
if p+1==N:
if x==k:
ans+=1
else:
while j<=N-1:
if x==k:
ans+=1
j+=1
t+=1
if j!=N:
x=B[j]-B[t-1]
else:
if j+abs(x-k)<=N-1:
j+=abs(x-k)
t+=abs(x-k)
x=B[j]-B[t-1]
else:
break
k+=1
print(ans)``````

Birthday Gift Again CodeChef Solution in C

``````#include <stdio.h>
#include <string.h>

int main(){
int ts;
scanf("%d",&ts);
while (ts--)
{
char s[100010];
scanf("%s",s);
int Ans = 0;
int ln = strlen(s);
for(int i = 1; (i*i + i)<= ln;i++){
int k = i*i+i;
int front = 0,rear = 0;
long long one = 0,zero = 0;
for(int x = 0; x <ln;x++){
if(front<k-1) {
if(s[front++] =='1') one++;
}
else if(front ==k-1){
if(s[front++] =='1') one++;
zero  = k-one;
if(zero == one*one) {
Ans++;
}
}
else{
if(s[rear++] == '1') one--;
if(s[front++]=='1') one++;
zero  = k-one;
if(zero == one*one) {
Ans++;
}
}
}
}
printf("%d\n",Ans);
}
return 0;
}``````

Birthday Gift Again CodeChef Solution in JAVA

``````import java.io.*;
import java.util.*;

class Solution {

public static void main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
// sc.nextLine();
while(n-->0){
String s = sc.next();
int l = s.length();
int c = 1;
int ans=0;
int win = c + (c*c);
while (win<=l){
int cnt = count(s.substring(0,win));
if(cnt==c){
ans++;
}
int b=-1;

for (int i = win; i <l; i++) {
b++;
if(s.charAt(i)=='1'){
cnt++;
}
if(s.charAt(b)=='1')
cnt--;
if(cnt==c)
ans++;

}
c++;
win = c + (c*c);

}
System.out.println(ans);

}
}
static int count(String s){
int res =0;
for (int i = 0; i <s.length() ; i++) {
if(s.charAt(i)=='1'){
res++;
}
}
return res;
}

}``````

Birthday Gift Again CodeChef Solution in PYPY 3

``````for _ in range(int(input())):
data = list(map(int,input()))
n = len(data)

# Represents count of ones in data[0:i]
pre = [0] * (n+1)
for i in range(n):
pre[i+1]=pre[i]+data[i]

total = 0
for k in range(1,n):
# window needs k*k ones and k zeroes
width = k*k + k - 1
if width>n:
break

for l in range(1,n+1 - width):
r = l + width
ones = pre[r] - pre[l-1]
#            zeroes = r-(l-1) - ones
#            print(l,r,ones,zeroes)
if ones == k:
total += 1

print(total)``````
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