sicily 1001. Alphacode dp

看了动态规划，对dp有个大概理解，这题之前一直卡着，刚开始还用搜索做~

要注意对0的处理，开始把方程列错，WA数次

dp方程为 a
= a[n-1]+a[n-2]，输入的字符串一次为X1X2X3…Xn…

当Xn = 0时，a
= a[n-2]

当Xn != 0时，

若X(n-1)与X(n)组成的数大于26， a
= a[n-1].

若X(n-1)与X(n)组成的数<=26,

x(n-1)不为0时, a
= a[n-1] + a[n-2]

x(n-1) = 0时， a
= a[n-1]

#include <iostream>
#include <string>
#define MAX 100000
using namespace std;

int main()
{
string num;
long long a[MAX];
int size;
int n;
while (cin >> num && num != "0")
{
size = num.length();
a[0] = 1;
a[1] = 1;

if (size == 1)
{
cout << "1" << endl;
continue;
}

//对第二位处理
n = (num[0]-'0')*10 + num[1]- '0';
if (n <= 26 && num[1] != '0')
a[1] = 2;

for (int i = 2; i < size; i++)
{
if (num[i] == '0')
a[i] = a[i-2];
else
{
a[i] = a[i-1];
n = (num[i-1]-'0')*10 + num[i]-'0';

if (n <= 26 && num[i-1] != '0')
a[i] += a[i-2];
}

}
cout << a[size-1] << endl;
}

return 0;
}