523

题目

Description

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 500 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 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 long variable.

25114
1111111111
3333333333
0

6
89
1

题解

dp[i] 表示到第 i 个字符的可能个数

• 如果第 i 个数可以和第 i-1 个数连起来构成一个合法的数( 1~26 )
那么,它有两种计算情况:

• 不和前面的数连起来,将他单独看作一个数,有 dp[i-1] 种可能

• 和前面的数连起来,共同构成一个数
这与上一个数( i-1 )单独看作一个数的数量一样,即 dp[i-2]

• 如果第 i 个数不能和第 i-1 个数连起来构成一个合法的数
那么,显然只有一种情况,就是将第 i 个数,单独看成一个,也即 dp[i] = dp[i-1]

代码

/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！

*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <functional>
using namespace std;

const int maxn = 50005;
char s[maxn];
int dp[maxn];
int ss[maxn];

bool Do() {
scanf("%s",s);
if(s[0] == '0')
return false;
int len = strlen(s);

int pos = 1;
for(int i = 0;i < len;i++) {
ss[pos] = s[i] - '0';
if(ss[pos] == 0) {
ss[pos - 1] = ss[pos - 1] * 10 + ss[pos];
} else {
pos++;
}
}

memset(dp,0,sizeof(dp));
dp[0] = dp[1] = 1;

for(int i = 2;i < pos;i++) {
int num = ss[i - 1];
int k = ss[i];
while(k) {
num *= 10;
k /= 10;
}
num += (ss[i]);

if(num <= 26)
dp[i] = dp[i - 1] + dp[i - 2];
else
dp[i] = dp[i - 1];

}

printf("%d\n",dp[pos - 1]);

return true;
}

int main() {
while(Do());
return 0;
}

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