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A-E Hash Function CodeChef Solution

A-E Hash Function CodeChef Solution in C++17

``````#include <bits/stdc++.h>

#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("Ofast")//Comment optimisations for interactive problems (use endl)
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")

using namespace std;

struct PairHash {inline std::size_t operator()(const std::pair<int, int> &v) const { return v.first * 31 + v.second; }};

// speed
#define Code ios_base::sync_with_stdio(false);
#define By ios::sync_with_stdio(0);
#define Sumfi cout.tie(NULL);

// alias
using ll = long long;
using ld = long double;
using ull = unsigned long long;

// constants
const ld PI = 3.14159265358979323846;  /* pi */
const ll INF = 1e18;
const ld EPS = 1e-9;
const ll MAX_N = 202020;
const ll mod = 1000000007;

// typedef
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef array<ll,3> all3;
typedef array<ll,5> all5;
typedef vector<all3> vall3;
typedef vector<all5> vall5;
typedef vector<ld> vld;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<int> vi;
typedef deque<ll> dqll;
typedef deque<pll> dqpll;
typedef pair<string, string> pss;
typedef vector<pss> vpss;
typedef vector<string> vs;
typedef vector<vs> vvs;
typedef unordered_set<ll> usll;
typedef unordered_set<pll, PairHash> uspll;
typedef unordered_map<ll, ll> umll;
typedef unordered_map<pll, ll, PairHash> umpll;

// macros
#define rep(i,m,n) for(ll i=m;i<n;i++)
#define rrep(i,m,n) for(ll i=n;i>=m;i--)
#define all(a) begin(a), end(a)
#define rall(a) rbegin(a), rend(a)
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define INF(a) memset(a,0x3f3f3f3f3f3f3f3fLL,sizeof(a))
#define ASCEND(a) iota(all(a),0)
#define sz(x) ll((x).size())
#define BIT(a,i) (a & (1ll<<i))
#define BITSHIFT(a,i,n) (((a<<i) & ((1ll<<n) - 1)) | (a>>(n-i)))
#define pyes cout<<"YES\n";
#define pno cout<<"NO\n";
#define endl "\n"
#define pneg1 cout<<"-1\n";
#define ppossible cout<<"Possible\n";
#define pimpossible cout<<"Impossible\n";
#define TC(x) cout<<"Case #"<<x<<": ";
#define X first
#define Y second

// utility functions
template <typename T>
void print(T &&t)  { cout << t << "\n"; }
template<typename T>
void printv(vector<T>v){ll n=v.size();rep(i,0,n){cout<<v[i];if(i+1!=n)cout<<' ';}cout<<endl;}
template<typename T>
void printvln(vector<T>v){ll n=v.size();rep(i,0,n)cout<<v[i]<<endl;}
void fileIO(string in = "input.txt", string out = "output.txt") {freopen(in.c_str(),"r",stdin); freopen(out.c_str(),"w",stdout);}
template<typename T>
template<typename T, typename U>
template<typename T, typename U>

struct Combination {
vll fac, inv;
ll n, MOD;

ll modpow(ll n, ll x, ll MOD = mod) { if(!x) return 1; ll res = modpow(n,x>>1,MOD); res = (res * res) % MOD; if(x&1) res = (res * n) % MOD; return res; }

Combination(ll _n, ll MOD = mod): n(_n + 1), MOD(MOD) {
inv = fac = vll(n,1);
rep(i,1,n) fac[i] = fac[i-1] * i % MOD;
inv[n - 1] = modpow(fac[n - 1], MOD - 2, MOD);
rrep(i,1,n - 2) inv[i] = inv[i + 1] * (i + 1) % MOD;
}

ll fact(ll n) {return fac[n];}
ll nCr(ll n, ll r) {
if(n < r or n < 0 or r < 0) return 0;
return fac[n] * inv[r] % MOD * inv[n-r] % MOD;
}
};

struct Matrix {
ll r,c;
vvll matrix;
Matrix(ll r, ll c, ll v = 0): r(r), c(c), matrix(vvll(r,vll(c,v))) {}

Matrix operator*(const Matrix& B) const {
Matrix res(r, B.c);
rep(i,0,r) rep(j,0,B.c) rep(k,0,B.r) {
res.matrix[i][j] = (res.matrix[i][j] + matrix[i][k] * B.matrix[k][j] % mod) % mod;
}
return res;
}

Matrix copy() {
Matrix copy(r,c);
copy.matrix = matrix;
return copy;
}

Matrix pow(ll n) {
assert(r == c);
Matrix res(r,r);
Matrix now = copy();
rep(i,0,r) res.matrix[i][i] = 1;
while(n) {
if(n & 1) res = res * now;
now = now * now;
n /= 2;
}
return res;
}
};

// geometry data structures
template <typename T>
struct Point {
T y,x;
Point(T y, T x) : y(y), x(x) {}
Point(pair<T,T> p) : y(p.first), x(p.second) {}
Point() {}
void input() {cin>>y>>x;}
friend ostream& operator<<(ostream& os, const Point<T>& p) { os<<p.y<<' '<<p.x<<'\n'; return os;}
Point<T> operator+(Point<T>& p) {return Point<T>(y + p.y, x + p.x);}
Point<T> operator-(Point<T>& p) {return Point<T>(y - p.y, x - p.x);}
Point<T> operator*(ll n) {return Point<T>(y*n,x*n); }
Point<T> operator/(ll n) {return Point<T>(y/n,x/n); }

Point<T> rotate(Point<T> center, ld angle) {
ld si = sin(angle * PI / 180.), co = cos(angle * PI / 180.);
ld y = this->y - center.y;
ld x = this->x - center.x;

return Point<T>(y * co - x * si + center.y, y * si + x * co + center.x);
}
ld distance(Point<T> other) {
T dy = abs(this->y - other.y);
T dx = abs(this->x - other.x);
return sqrt(dy * dy + dx * dx);
}
};

template<typename T>
struct Line {
Point<T> A, B;
Line(Point<T> A, Point<T> B) : A(A), B(B) {}
Line() {}

void input() {
A = Point<T>();
B = Point<T>();
A.input();
B.input();
}

T ccw(Point<T> &a, Point<T> &b, Point<T> &c) {
T res = a.x * b.y + b.x * c.y + c.x * a.y;
res -= (a.x * c.y + b.x * a.y + c.x * b.y);
return res;
}

bool isIntersect(Line<T> o) {
T p1p2 = ccw(A,B,o.A) * ccw(A,B,o.B);
T p3p4 = ccw(o.A,o.B,A) * ccw(o.A,o.B,B);
if (p1p2 == 0 && p3p4 == 0) {
pair<T,T> p1(A.y, A.x), p2(B.y,B.x), p3(o.A.y, o.A.x), p4(o.B.y, o.B.x);
if (p1 > p2) swap(p2, p1);
if (p3 > p4) swap(p3, p4);
return p3 <= p2 && p1 <= p4;
}
return p1p2 <= 0 && p3p4 <= 0;
}

pair<bool,Point<ld>> intersection(Line<T> o) {
if(!this->intersection(o)) return {false, {}};
ld det = 1. * (o.B.y-o.A.y)*(B.x-A.x) - 1.*(o.B.x-o.A.x)*(B.y-A.y);
ld t = ((o.B.x-o.A.x)*(A.y-o.A.y) - (o.B.y-o.A.y)*(A.x-o.A.x)) / det;
return {true, {A.y + 1. * t * (B.y - A.y), B.x + 1. * t * (B.x - A.x)}};
}

//@formula for : y = ax + b
//@return {a,b};
pair<ld, ld> formula() {
T y1 = A.y, y2 = B.y;
T x1 = A.x, x2 = B.x;
if(y1 == y2) return {1e9, 0};
if(x1 == x2) return {0, 1e9};
ld a = 1. * (y2 - y1) / (x2 - x1);
ld b = -x1 * a + y1;
return {a, b};
}
};

template<typename T>
struct Circle {
Point<T> center;
Circle() {}

void input() {
center = Point<T>();
center.input();
}

bool circumference(Point<T> p) {
return (center.x - p.x) * (center.x - p.x) + (center.y - p.y) * (center.y - p.y) == radius * radius;
}

bool intersect(Circle<T> c) {
T d = (center.x - c.center.x) * (center.x - c.center.x) + (center.y - c.center.y) * (center.y - c.center.y);
}

bool include(Circle<T> c) {
T d = (center.x - c.center.x) * (center.x - c.center.x) + (center.y - c.center.y) * (center.y - c.center.y);
}
};

ll __gcd(ll x, ll y) { return !y ? x : __gcd(y, x % y); }
all3 __exgcd(ll x, ll y) { if(!y) return {x,1,0}; auto [g,x1,y1] = __exgcd(y, x % y); return {g, y1, x1 - (x/y) * y1}; }
ll __lcm(ll x, ll y) { return x / __gcd(x,y) * y; }
ll modpow(ll n, ll x, ll MOD = mod) { n%=MOD; if(!x) return 1; ll res = modpow(n,x>>1,MOD); res = (res * res) % MOD; if(x&1) res = (res * n) % MOD; return res; }

umll dp[55][55];

void helper(ll a, ll e) {
if(a == 0) dp[a][e] = {{0,1}};
if(sz(dp[a][e])) return;
ll n = a + e, l = n / 2, r = n - l;
umll& now = dp[a][e];
rep(i,0,min(a,l) + 1) {
ll A = i, E = l - i;
if(E > e) continue;
helper(A,E);
helper(a-A,e-E);
for(auto& [k1,v1] : dp[A][E]) {
for(auto& [k2,v2] : dp[a-A][e-E]) {
ll score = max(k1,k2) + a;
if(score > 1000) continue;
now[score] = (now[score] + v1 * v2 % mod) % mod;
}
}
}
}

ll solve(ll a, ll e, ll v) {
dp[0][0] = {{0,1}};
dp[1][0] = {{1,1}};
helper(a,e);
if(!dp[a][e].count(v)) return 0;
return dp[a][e][v];
}

int main() {
Code By Sumfi
cout.precision(12);
ll tc = 1;
cin>>tc;
rep(i,1,tc+1) {
ll a,e,v;
cin>>a>>e>>v;
print(solve(a,e,v));
}
return 0;
}``````

A-E Hash Function CodeChef Solution in C++14

``````#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
int DP[60][60][200];
const long long MOD = 1000000007;
long long funct(int a, int e, int v){if (v < 0){return 0;} if (DP[a][e][v] != -1){return DP[a][e][v];}
if (a+e <= 1){return DP[a][e][v] = (a<=v);}
long long sol = 0LL; int t = (a+e)/2; for (int i = 0;  i < t+1; i++)
{if ((i <= a) && (t-i <= e)){sol += funct(i,t-i,v-a)*funct(a-i,e-(t-i),v-a); sol%=MOD;}}
return DP[a][e][v] = sol;}
int main(){int t; memset(DP,-1,sizeof(DP)); scanf("%d",&t);
while (t--){int A, E, V; scanf ("%d%d%d", &A, &E, &V); long long temp = 0;
temp = (funct(A,E,V)-funct(A,E,V-1)+MOD)%MOD; printf("%lld\n",temp);}}``````

A-E Hash Function CodeChef Solution in C

``````#include<stdio.h>
#include<string.h>
#define ll long long
#define M 1000000007

ll dp[60][60][160];

ll S(int a,int e,int v){
int m=(a+e)/2,i;
ll p=0;

if(v<0||~dp[a][e][v]||a+e<=1)
return(v<0)?0:((~dp[a][e][v])?dp[a][e][v]:(dp[a][e][v]=(a<=v)));

for(i=-1;++i<=m;p=(i<=a&&m-i<=e)?(p+S(i,m-i,v-a)*S(a-i,e-(m-i),v-a))%M:p);

return dp[a][e][v]=p;
}

main()
{
int f,a,e,v;
ll r;
for(scanf("%d",&f) && memset(dp,-1,sizeof dp);
f--;
scanf("%d%d%d",&a,&e,&v),printf("%lld\n",(r=v>152?0:(r=(S(a,e,v)-S(a,e,v-1))%M)<0?r+M:r)));
return 0;
}
``````

A-E Hash Function CodeChef Solution in JAVA

``````import java.io.DataInputStream;
import java.io.InputStream;

public class Main{
static long hashedcounts[][][];
static int vlist[][][];
static int vlistindex[][];
static int mod=1000000007;
static int reqs[][];
public static void main(String args[]) throws Exception
{
ParserCount p=new ParserCount(System.in);
hashedcounts=new long[101][51][1001];
vlist=new int[101][51][1001];
vlistindex=new int[101][51];
int T=p.nextInt();
reqs=new int[101][51];
int inputs[][]=new int[T][3];
StringBuffer sb=new StringBuffer();
for(int i=0;i<T;i++)
{
int A=p.nextInt();int E=p.nextInt();int V=p.nextInt();
inputs[i][0]=A;inputs[i][1]=E;inputs[i][2]=V;
if((A+E)>50)
{
reqs[A+E][A]=1;
}
}
preProcess();
populate();
int A,E,V;
for(int i=0;i<T;i++)
{
A=inputs[i][0];E=inputs[i][1];V=inputs[i][2];
int l=A+E;
//System.out.println(hashedcounts[l][A][V]);
sb.append(hashedcounts[l][A][V]+"\n");
}
System.out.print(sb.toString());
}
public static void populate()
{
for(int length=3;length<=100;length++)
{
//System.out.println("Starting L="+length+"...");
int leftl,rightl;
if(length%2==1)
{
leftl=length/2;
rightl=leftl+1;
}
else
{
leftl=length/2;
rightl=length/2;
}
for(int A=0;A<=length && A<=50;A++)
{
if(length>50)
{
if(reqs[length][A]==0)//these are not necessary
continue;
}
//System.out.println("\t A="+A);
int ways=A;//at root it has this many A's
for(int leftA=0;leftA<=A;leftA++)
{
int rightA=A-leftA;
//					if(leftA>leftl || rightA>rightl)
//						continue;
int lcount=vlistindex[leftl][leftA];
int rcount=vlistindex[rightl][rightA];
//cross product start
for(int m=0;m<lcount;m++)
{
for(int n=0;n<rcount;n++)
{
int val1=vlist[leftl][leftA][m];
int val2=vlist[rightl][rightA][n];
if(ways+val1>1000 || ways+val2>1000)//to be verified
continue;
if(val1>val2)
{
{
//System.out.println(vlistindex[length][A]);
vlist[length][A][vlistindex[length][A]]=ways+val1;
vlistindex[length][A]++;
}
hashedcounts[length][A][ways+val1]+=((hashedcounts[leftl][leftA][val1]*hashedcounts[rightl][rightA][val2])%mod);
hashedcounts[length][A][ways+val1]=hashedcounts[length][A][ways+val1]%mod;
}
else
{
{
vlist[length][A][vlistindex[length][A]]=ways+val2;
vlistindex[length][A]++;
}
hashedcounts[length][A][ways+val2]+=((hashedcounts[leftl][leftA][val1]*hashedcounts[rightl][rightA][val2])%mod);
hashedcounts[length][A][ways+val2]=hashedcounts[length][A][ways+val2]%mod;
}
}
}
}
//print stuff
//				for(int k=0;k<vlistindex[length][A];k++)
//				{
//					System.out.print("("+vlist[length][A][k]+" "+hashedcounts[length][A][vlist[length][A][k]]+")");
//				}
//System.out.println("\n");
}
}
}
public static void preProcess()
{
hashedcounts[0][0][0]=1;vlist[0][0][vlistindex[0][0]++]=0;
hashedcounts[1][0][0]=1;vlist[1][0][vlistindex[1][0]++]=0;
hashedcounts[1][1][1]=1;vlist[1][1][vlistindex[1][1]++]=1;
hashedcounts[2][0][0]=1;vlist[2][0][vlistindex[2][0]++]=0;
hashedcounts[2][1][2]=2;vlist[2][1][vlistindex[2][1]++]=2;
hashedcounts[2][2][3]=1;vlist[2][2][vlistindex[2][2]++]=3;
}
}
class ParserCount
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
public ParserCount(InputStream in)
{
din = new DataInputStream(in);
buffer = new byte[BUFFER_SIZE];
}
public String nextString() throws Exception
{
StringBuffer sb=new StringBuffer("");
while (c <= ' ') c = read();
do
{
sb.append((char)c);
}while(c!='\r');
return sb.toString();
}

public int nextInt() throws Exception
{
int ret = 0;
while (c <= ' ') c = read();
boolean neg = c == '-';
do
{
ret = ret * 10 + c - '0';
} while (c > ' ');
if (neg) return -ret;
return ret;
}

private void fillBuffer() throws Exception
{
if (bytesRead == -1) buffer[0] = -1;
}

{
return buffer[bufferPointer++];
}
}``````

A-E Hash Function CodeChef Solution in PYTH

``````import sys
import math

M = 1000000007

rs = {}

def fn(An, En, V):
global rs, M

try:
return rs[An][En][V]
except:
##        if V < 0:
##            return 0
if rs.has_key(An):
if not rs[An].has_key(En):
rs[An][En] = {}
else:
rs[An] = {}
rs[An][En] = {}

rs[An][En][V] = None
if V < An:
rs[An][En][V] = 0
else :
if An == 0:
rs[An][En][V] = 1
else:
if An == 1 and En == 0:
if V == 0:
rs[An][En][V] = 0
else:
rs[An][En][V] = 1

if rs[An][En][V] == None:
vRemain = V - An
s = (An + En) / 2
cnt = 0
for an1 in range(max(0, s - En), min(An, s)+1):
cnt += fn(an1, s - an1, vRemain) * fn(An - an1, En - s + an1, vRemain)
rs[An][En][V] = cnt % M
return rs[An][En][V]

def main():
cases = []
for n in range(0, testCases):
An, En, V = sys.stdin.readline().split(' ')
print "%s" % ((fn(int(An), int(En), int(V)) - fn(int(An), int(En), int(V)-1) + M ) % M)

if __name__ == '__main__':
main()``````

A-E Hash Function CodeChef Solution in C#

``````using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ProgrammingContest.CodeChef.Contest.July_2011_Long_Contest
{
class A_E_Hash_Function
{
const long mod = 1000000007;

static Dictionary<int, long>[,] memo = new Dictionary<int, long>[51, 51];

static Dictionary<int,long> rec(int A, int E)
{
if (memo[A, E] != null)
return memo[A,E];

memo[A, E] = new Dictionary<int, long>();
Dictionary<int, long> res = memo[A, E];
if (A + E <= 1 || A == 0)
{
res[A] = 1;
return res;
}

int n = (A + E) / 2;
for (int i = 0; i <= n; i++)
{
if (A < i || E < n - i)
continue;

var d1 = rec(i, n - i);
var d2 = rec(A - i, E - (n - i));

foreach (var v in d1)
foreach (var u in d2)
{
int max = Math.Max(v.Key, u.Key),
val = A + max;
if (!res.ContainsKey(val))
res[val] = 0;
res[val] = (res[val] + u.Value * v.Value) % mod;
}
}

return res;
}

static void Main()
{
while(T-- > 0)
{
var xs = Array.ConvertAll(Console.ReadLine().Split(' '), sss => int.Parse(sss));
int A = xs[0],
E = xs[1],
V = xs[2];
var dic = rec(A, E);
Console.WriteLine(dic.ContainsKey(V) ? dic[V] : 0);
}
}
}
}``````

A-E Hash Function CodeChef Solution in GO

``````package main

import (
"fmt"
)

const (
M = int32(1000000007)
M64 = int64(1000000007)
aMax = 50
eMax = 50
vMax = 1000
)

var rs [aMax + 1][eMax +1][vMax +1]int32

func fn(an, en, v int) int32{
if v < 0{
return 0
}
if rs[an][en][v] == -1{
if v < an {
rs[an][en][v] = 0
} else if an == 0{
rs[an][en][v] = 1
} else if an == 1 && en == 0{
if v == 0{
rs[an][en][v] = 0
} else {
rs[an][en][v] = 1
}
}
if rs[an][en][v] == -1{
r := v - an
s := (an + en) / 2
var cnt int64
start := s - en
if en > s{
start = 0
}
end := s
if an < s{
end = an
}
for a1 := start; a1 <= end ;a1++{
cnt = (cnt + (int64(fn(a1, s -a1, r)) * int64(fn(an-a1, en-s+a1, r))) ) % M64
}
rs[an][en][v] = int32(cnt)
}
}
return rs[an][en][v]
}

func main() {
var tn int
var out int32
for i :=0; i <= aMax; i++{
for j :=0; j <= eMax; j++{
for k :=0; k <= vMax; k++{
rs[i][j][k] = -1
}
}
}
fmt.Scanln(&tn)
for i :=0; i < tn; i++{
var a, e, v int
fmt.Scanln(&a, &e, &v)
out = (fn(a, e, v) - fn(a, e, v -1 ) + M) % M
fmt.Printf("%d\n", out)
}
}``````
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