POJ 1850 Code 統計問題

題意：存在下面的編碼方式：

a - 1
b - 2
...
z - 26
ab - 27
...
az - 51
bc - 52
...
vwxyz - 83681

其中字串的長度逐漸增加，並且每一個字串的字元只能是升序。例如b不能排在a的前面。

#include<cstdio>#include<cstring>using namespace std;#define lint __int64lint C ( int m, int n ){    if ( m < n ) return 0;    if ( m == n || n == 0 ) return 1;    lint sum = 1;    for ( int i = 1, j = m; i <= n; i++, j-- )        sum = sum * j / i;    return sum;}int main(){    char str[20];    scanf("%s",str);    int i, j;    for ( i = 1; str[i]; i++ )        if ( str[i] < str[i-1] )        {            printf("0\n"); return 0;        }    int len = strlen(str);    lint ret = 0;    for ( i = 1; i < len; i++ )        ret += C(26,i);    for ( i = 'a'; i < str[0]; i++ )        ret += C('z'-i,len-1);    for ( i = 1; i < len; i++ )        for ( j = str[i-1]+1; j < str[i]; j++ )            ret += C('z'-j,len-i-1);    printf("%I64d\n",ret+1);    return 0;}