Algorithm
Problem link- https://www.spoj.com/problems/ACODE/
ACODE - Alphacode
Alice and Bob need to send secret messages to each other and are discussing ways to encode their messages:
Alice: “Let’s just use a very simple code: We’ll assign ‘A’ the code word 1, ‘B’ will be 2, and so on down to ‘Z’ being assigned 26.”
Bob: “That’s a stupid code, Alice. Suppose I send you the word ‘BEAN’ encoded as 25114. You could decode that in many different ways!”
Alice: “Sure you could, but what words would you get? Other than ‘BEAN’, you’d get ‘BEAAD’, ‘YAAD’, ‘YAN’, ‘YKD’ and ‘BEKD’. I think you would be able to figure out the correct decoding. And why would you send me the word ‘BEAN’ anyway?”
Bob: “OK, maybe that’s a bad example, but I bet you that if you got a string of length 5000 there would be tons of different decodings and with that many you would find at least two different ones that would make sense.”
Alice: “How many different decodings?”
Bob: “Jillions!”
For some reason, Alice is still unconvinced by Bob’s argument, so she requires a program that will determine how many decodings there can be for a given string using her code.
Input
Input will consist of multiple input sets. Each set will consist of a single line of at most 5000 digits representing a valid encryption (for example, no line will begin with a 0). There will be no spaces between the digits. An input line of ‘0’ will terminate the input and should not be processed.
Output
For each input set, output the number of possible decodings for the input string. All answers will be within the range of a 64 bit signed integer.
Example
Input: 25114 1111111111 3333333333 0 Output: 6 89 1
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <bits/stdc++.h>
#include <limits.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<int> vi;
#define MOD (ll)1000000007
#define pb push_back
#define EPS 1e-9
#define FOR(i,n) for(int i = 0;i < n; i++)
#define FORE(i,a,b) for(int i = a;i <= b; i++)
#define pi(a) printf("%d\n", a)
#define all(c) c.begin(), c.end()
#define tr(container, it) for(typeof(container.begin()) it = container.begin(); it != container.end(); it++)
#define gc getchar_unlocked
#define sdi(a, b) si(a);si(b)
#define io ios_base::sync_with_stdio(false);cin.tie(NULL);
#define endl '\n'
#define F first
#define S second
template <typename T> T gcd(T a, T b){return (b==0)?a:gcd(b,a%b);}
template <typename T> T lcm(T a, T b){return a*(b/gcd(a,b));}
template <typename T> T mod_exp(T b, T p, T m){T x = 1;while(p){if(p&1)x=(x*b)%m;b=(b*b)%m;p=p>>1;}return x;}
template <typename T> T invFermat(T a, T p){return mod_exp(a, p-2, p);}
template <typename T> T exp(T b, T p){T x = 1;while(p){if(p&1)x=(x*b);b=(b*b);p=p>>1;}return x;}
void si(int &x){
register int c = gc();
x = 0;
int neg = 0;
for(;((c<48 || c>57) && c != '-');c = gc());
if(c=='-') {neg=1;c=gc();}
for(;c>47 && c<58;c = gc()) {x = (x<<1) + (x<<3) + c - 48;}
if(neg) x=-x;
}
int main(){
io;
string input;
while(1){
cin >> input;
if(input == "0")
break;
ll dp[input.size()+1];
memset(dp, 0, sizeof(dp));
//dp[i] represents number of decodings of first i characters
dp[0] = 1;
dp[1] = 1;
for(int i = 2;i <= input.size(); i++){
int current = input[i-1]-'0';
int previous = input[i-2]-'0';
if(current)
dp[i] = dp[i-1];
int num = previous*10+current;
if(num >= 10 && num <= 26){
dp[i] += dp[i-2];
}
}
cout << dp[input.size()] << endl;
}
return 0;
}Input
1111111111
3333333333
0
Output
89
1
#2 Code Example with Java Programming
Code -
Java Programming
import java.io.*;
class alphacode{
public static int ans;
public static void main(String[] args) throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String s=br.readLine();
int x=Integer.parseInt(Character.toString(s.charAt(0)));
while(x!=0){
ans=0;
int a[]=new int[s.length()+1];
for(int i=1;i<=s.length();i++){
a[i]=Integer.parseInt(Character.toString(s.charAt(i-1)));
}
// function(a,1,a.length);
long dp[]=new long[a.length+1];
dp[0]=1;
dp[1]=1;
int n=s.length()+1;
for(int i=2;i<n;i++){
if(a[i]==0){
dp[i]=dp[i-2];
}else if(a[i]<=6){
if(a[i-1]==1 || a[i-1]==2){
dp[i]=dp[i-1]+dp[i-2];
}else{
dp[i]=dp[i-1];
}
}else{
if(a[i-1]==1 || a[i-1]==2){
dp[i]=dp[i-1]+dp[i-2];
}else{
dp[i]=dp[i-1];
}
}
// System.out.println(a[i]+" "+dp[i]);
}
// ans=dp[a.length-1];
System.out.println(dp[a.length-1]);
s=br.readLine();
x=Integer.parseInt(Character.toString(s.charAt(0)));
}
}
public static void function(int a[],int i,int n){
if(i>=n){
ans+=1;
return;
}else{
if(a[i]==1){
if(i<n-1){
if(a[i+1]==0){
function(a,i+1,n);
}else{
function(a,i+1,n);
function(a,i+2,n);
}
}else{
function(a,i+1,n);
}
}else if(a[i]==2){
function(a,i+1,n);
function(a,i+2,n);
}else{
function(a,i+1,n);
}
}
}
}Input
1111111111
3333333333
0
Output
89
1
Demonstration
SPOJ Solution-Alphacode-Solution in C, C++, Java, Python