# PROBLEM LINK:CodeChef: Practical coding for everyone

Practice

Author: targer_5star
Tester: targer_5star
Editorialist: targer_5star

EASY

Math etc.

# PROBLEM:

Analyze the Pattern of the following series Which is reducing in a particular way and Find the Nth term of the series.

500,499,498,490,486,459…….

Find and Print The Nth Term of the series

# QUICK EXPLANATION:

The pattern is as follows:- 500,500-1^3,500-1^2,500-2^3,500-2^2,500-3^3,500-3^2… so on up to 0.

# EXPLANATION:

The pattern is as follows:- (500), (500-1^3), (500-1^2), (500-2^3), (500-2^2), (500-3^3), (500-3^2),…,(0) so on up to 0.

# SOLUTIONS:

Setter's Solution

#include <bits/stdc++.h>

const long long SZ = 4e3 + 7;

const long long inf = 1e18;

const long long MOD = 1e9 + 7;

const long long mod = 1e9 + 7;

long long opnmbr = 1;

#define ll long long

#define ld long double

#define pb push_back

#define mp make_pair

#define eb emplace_back

#define pll pair<ll, ll>

#define vi vector

#define vs vector

#define vpl vector

#define qi queue

#define si set

#define map map<ll, ll>

#define umap unordered_map<ll, ll>

#define fi first

#define se second

#define sz(x) (ll)x.size()

#define all(c) (c).begin(), (c).end()

#define allr(c) (c).rbegin(), (c).rend()

#define Max(a,b) ((a > b) ? a : b)

#define Min(a,b) ((a < b) ? a : b)

#define ci(X) ll X; cin>>X

#define cii(X, Y) ll X, Y; cin>>X>>Y

#define ciii(X, Y, Z) ll X, Y, Z; cin>>X>>Y>>Z

#define ciiii(W, X, Y, Z) ll W, X, Y, Z; cin>>W>>X>>Y>>Z

#define co cout<<

#define in cin>>

#define Jivan_ka_asli_aadhar_to_prem_hai_na ll ___T; cin>>___T; while (___T-- > 0)

#define Code_Wode_mai_kya_rakha_hai ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)

#define inf 1e18

#define ed endl

#define abhi(i, n) for(ll i = 0; i < (n); i++)

#define abhi1(i, n) for(ll i = 1; i <= (n); i++)

#define abhi2(i, l, r) for(ll i = l; i < (r); i++)

#define rabhi(i, l, r) for(ll i = l; i >= (r); i–)

#define ms0(X) memset((X), 0, sizeof((X)))

#define ms1(X, V) memset((X), 1, sizeof((X)))

#define ms2(X, V) memset((X), 1, sizeof((X)))

#define flv(X, V) fill(all((X)), V)

#define gcd(a,b) __gcd(a,b)

using namespace std;

ll prmod(ll a, ll b)

{

``````ll ans = 1;

while (b)

{

if (b & 1) ans = (ans * a) % MOD;

b = b / 2;

a = (a * a) % MOD;

}

return ans;
``````

}

int pri(ll n)

{

int flag=0;

for(ll i = 2; i <= n / 2; ++i)

{

`````` if(n % i == 0)

{

flag = 1;

break;

}
``````

}

if(flag==0)

{

``````  return 1;
``````

}

else

{

``````   return 0;
``````

}

}

ll lcm(ll a, ll b)

{

`````` return (a / gcd(a, b)) * b;
``````

}

{

int n,a=500;

cin>>n;

if(n==1)

{

``````   cout<<500<<"\n";
``````

}

else

{

``````  int j=1;

for(int i=2;i<=n;i=i+2)

{

a=a-(j*j*j);

if(i==n)

{

cout<<a<<"\n";

}

a=a-(j*j);

if(i+1==n)

{

cout<<a<<"\n";

}

j++;

}
``````

}

}

int main()

{

``````   Code_Wode_mai_kya_rakha_hai;

{

}
``````

}

Tester's Solution

{

int n,a=500;

cin>>n;

if(n==1)

{

``````   cout<<500<<"\n";
``````

}

else

{

``````  int j=1;

for(int i=2;i<=n;i=i+2)

{

a=a-(j*j*j);

if(i==n)

{

cout<<a<<"\n";

}

a=a-(j*j);

if(i+1==n)

{

cout<<a<<"\n";

}

j++;

}
``````

}

}

Editorialist's Solution

#include <bits/stdc++.h>

const long long SZ = 4e3 + 7;

const long long inf = 1e18;

const long long MOD = 1e9 + 7;

const long long mod = 1e9 + 7;

long long opnmbr = 1;

#define ll long long

#define ld long double

#define pb push_back

#define mp make_pair

#define eb emplace_back

#define pll pair<ll, ll>

#define vi vector

#define vs vector

#define vpl vector

#define qi queue

#define si set

#define map map<ll, ll>

#define umap unordered_map<ll, ll>

#define fi first

#define se second

#define sz(x) (ll)x.size()

#define all(c) (c).begin(), (c).end()

#define allr(c) (c).rbegin(), (c).rend()

#define Max(a,b) ((a > b) ? a : b)

#define Min(a,b) ((a < b) ? a : b)

#define ci(X) ll X; cin>>X

#define cii(X, Y) ll X, Y; cin>>X>>Y

#define ciii(X, Y, Z) ll X, Y, Z; cin>>X>>Y>>Z

#define ciiii(W, X, Y, Z) ll W, X, Y, Z; cin>>W>>X>>Y>>Z

#define co cout<<

#define in cin>>

#define Jivan_ka_asli_aadhar_to_prem_hai_na ll ___T; cin>>___T; while (___T-- > 0)

#define Code_Wode_mai_kya_rakha_hai ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)

#define inf 1e18

#define ed endl

#define abhi(i, n) for(ll i = 0; i < (n); i++)

#define abhi1(i, n) for(ll i = 1; i <= (n); i++)

#define abhi2(i, l, r) for(ll i = l; i < (r); i++)

#define rabhi(i, l, r) for(ll i = l; i >= (r); i–)

#define ms0(X) memset((X), 0, sizeof((X)))

#define ms1(X, V) memset((X), 1, sizeof((X)))

#define ms2(X, V) memset((X), 1, sizeof((X)))

#define flv(X, V) fill(all((X)), V)

#define gcd(a,b) __gcd(a,b)

using namespace std;

ll prmod(ll a, ll b)

{

``````ll ans = 1;

while (b)

{

if (b & 1) ans = (ans * a) % MOD;

b = b / 2;

a = (a * a) % MOD;

}

return ans;
``````

}

int pri(ll n)

{

int flag=0;

for(ll i = 2; i <= n / 2; ++i)

{

`````` if(n % i == 0)

{

flag = 1;

break;

}
``````

}

if(flag==0)

{

``````  return 1;
``````

}

else

{

``````   return 0;
``````

}

}

ll lcm(ll a, ll b)

{

`````` return (a / gcd(a, b)) * b;
``````

}

{

int n,a=500;

cin>>n;

if(n==1)

{

``````   cout<<500<<"\n";
``````

}

else

{

``````  int j=1;

for(int i=2;i<=n;i=i+2)

{

a=a-(j*j*j);

if(i==n)

{

cout<<a<<"\n";

}

a=a-(j*j);

if(i+1==n)

{

cout<<a<<"\n";

}

j++;

}
``````

}

}

int main()

{

``````   Code_Wode_mai_kya_rakha_hai;

{

}
``````

}

1 Like

17 posts were merged into an existing topic: Use this for anything related to plagiarism - Reporting cheating, appeals, etc

print("Series is ")
print(“500,499,498,490,486,459,…”)
a = int(input("Enter the term you want to print: "))
b = [500]
t = 500
for i in range(1,a//2+1):
for j in range(3,1,-1):
t = t - pow(i,j)
b.append(t)
print(b)
print(“The {}th term of this series is {}”.format(a,b[a-1]))

1 Like