Merge pull request ashutosh97#119 from Udit-8/master

Added question alphacode

Author: ashutosh97

alphacode/problem_statement.txt

@@ -0,0 +1,34 @@
+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

alphacode/solution.txt

@@ -0,0 +1,32 @@
+#include <cstdio>
+#include <cstring>
+
+using namespace std;
+
+int main(){
+	int L;
+	char d[5001];
+	long long dp[5001];
+	
+	while(true){
+		scanf("%s",d);
+		if(d[0]=='0') break;
+		
+		L = strlen(d);
+		
+		dp[0] = dp[1] = 1;
+		
+		for(int i = 2;i<=L;++i){
+			dp[i] = 0;
+			
+			char c1 = d[i-2]-'0', c2 = d[i-1]-'0';
+			
+			if(c1==1 || (c1==2 && c2<=6)) dp[i] += dp[i-2];
+			if(c2!=0) dp[i] += dp[i-1];
+		}
+		
+		printf("%lld\n",dp[L]);
+	}
+	
+	return 0;
+}