In this HackerEarth Soft Soft problem solution Let us define an easy Sorting Algorithm called SoftSort. SoftSort Sorting algorithm involves the use of IF and ELSE decision statements only. For example :
To sort two numbers a and b. SoftSort Algorithm's Code is given below.
void print(int a,int b){
printf( "%d %d\n",a,b);
}
void sort( int a, int b){
if( b < a )
print( b, a );
else
print( a, b );
}
To sort three numbers a , b and c . Source Code is given below.
void print( int a, int b, int c ) {
printf( "%d %d %d\n", a, b, c );
}
void sort( int a, int b, int c ) {
if( a < b )
if( a < c )
if( b < c )
print( a, b, c );
else
print( a, c, b );
else
print( c, a, b );
else
if( b < c )
if( a < c )
print( b, a, c );
else
print( b, c, a );
else
print( c, b, a );
}
ANI is fascinated with the SoftSort Algorithm and decided to ask an interesting question to KUKU.
What could be the length of source code to sort n numbers using SoftSort Algorithm?
HackerEarth Soft Sort problem solution.
#include<iostream>
#include<bits/stdc++.h>
using namespace std;
#define MOD 1000000007
#define ft first
#define sd second
#define VI vector<int>
#define VLL vector<long long int>
#define PII pair<int,int>
#define pb push_back
#define rsz(v,n) v.resize(n)
// input and output
#define scan(x) scanf("%d",&x)
#define scanll(x) scanf("%lld",&x)
#define ll long long int
#define rep(i,x,y) for(i=x;i<y;i++)
#define print(x) printf("%d\n",x)
#define printll(x) printf("%lld\n",x)
#define all(v) v.begin(),v.end()
#define ms(v) memset(v,0,sizeof(v))
#define FOR(i,a,b) for(i=a;i<b;i++)
#define PIE 3.14159265358979323846264338327950
#ifdef ONLINE_JUDGE
inline void inp( int &n )
{
n=0;
int ch=getchar_unlocked();int sign=1;
while( ch < '0' || ch > '9' ){if(ch=='-')sign=-1; ch=getchar_unlocked();}
while( ch >= '0' && ch <= '9' )
n = (n<<3)+(n<<1) + ch-'0', ch=getchar_unlocked();
n=n*sign;
}
#else
inline void inp(int &n){
cin>>n;
}
#endif
ll fact[1000001];
void pre_process()
{
fact[0]=1;
for(int i=1;i<=1000000;i++)
{
fact[i]=(fact[i-1]*i)%MOD;
}
}
int main()
{
int n,t;
pre_process();
inp(t);
assert(1<=t&&t<=100000);
while(t--)
{
int n;
inp(n);
assert(1<=n&&n<=1000000);
ll ans=((3*fact[n])%MOD+3)%MOD;
printll(ans);
}
return 0;
}
Second solution
#include <cstdlib>
#include <stdio.h>
#include <cstring>
#include <complex>
#include <vector>
#include <cmath>
#include <ctime>
#include <iostream>
#include <numeric>
#include <algorithm>
#include <map>
#include <utility>
#include <set>
#include <stack>
#include <queue>
#include <iomanip>
#include <locale>
#include <sstream>
#include <string>
#define FOR(i,n) for(i=0;i<n;i++)
#define FORI(i,a,n) for(i=a;i<n;i++)
#define FORC(it,C) for(it=C.begin();it!=C.end();it++)
#define scanI(x) scanf("%d",&x)
#define scanD(x) scanf("%lf",&x)
#define print(x) printf("%d\n",x)
#define MAX 2000004
#define MOD 1000000007
typedef long long ll;
using namespace std;
ll fact[MAX],ifact[MAX];
ll power(ll n,int m)
{
if(m==0) return 1;
ll x=power(n,m/2);
if(m%2==0)
return (x*x)%MOD;
else
return (((x*x)%MOD)*n)%MOD;
}
void preProcess()
{
fact[0]=1;
ifact[0]=1;
int i;
for(i=1;i<MAX;i++)
{
fact[i]=fact[i-1]*i;
fact[i]%=MOD;
}
ifact[2000000] = power(fact[2000000],MOD-2);
for(i=2000000;i>1;i--)
ifact[i-1] = ifact[i]*i%MOD;
}
ll comb(int a)
{
return (((fact[a]*3)%MOD+3)%MOD)%MOD;
}
int main() {
std::ios_base::sync_with_stdio(false);
int t;
cin >> t;
ll ans;
preProcess();
while(t--){
ll n;
cin >> n;
ans=comb(n);
ans%=MOD;
printf("%lld\n",ans);
}
return 0;
}
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