题目:
Transmitting and memorizing information is a task that requires different coding systems for the best use of the available space. A well known system is that one where a number is associated to a character sequence. It is considered that the words are made only of small characters of the English alphabet a,b,c, ..., z (26 characters). From all these words we consider only those whose letters are in lexigraphical order (each character is smaller than the next character).
The coding system works like this:
• The words are arranged in the increasing order of their length.
• The words with the same length are arranged in lexicographical order (the order from the dictionary).
• We codify these words by their numbering, starting with a, as follows:
a - 1
b - 2
...
z - 26
ab - 27
...
az - 51
bc - 52
...
vwxyz - 83681
...
Specify for a given word if it can be codified according to this coding system. For the affirmative case specify its code.
The coding system works like this:
• The words are arranged in the increasing order of their length.
• The words with the same length are arranged in lexicographical order (the order from the dictionary).
• We codify these words by their numbering, starting with a, as follows:
a - 1
b - 2
...
z - 26
ab - 27
...
az - 51
bc - 52
...
vwxyz - 83681
...
Specify for a given word if it can be codified according to this coding system. For the affirmative case specify its code.
Input
The only line contains a word. There are some constraints:
• The word is maximum 10 letters length
• The English alphabet has 26 characters.
• The word is maximum 10 letters length
• The English alphabet has 26 characters.
Output
The output will contain the code of the given word, or 0 if the word can not be codified.
Sample Input
bf
Sample Output
55
题意分析:
对于不同的小写字母,严格升序组成的一组序列,分别代表从1~n的数值。
这题关键是求组合的数值C。对于方法,并不是太难,分两种情况
1.当长度小于给出的字符串L时,对于字符串长度为1的字符串总共代表的数字的量就是C(26,1),对于长度为2的字符串总共代表的数字的量就是C(26,2),一次类推,直到C(26,L-1)。
2.对于长度等于L的,从最开头的字符s[0]开始,对于比这个字符严格小的,如果有一个,我们可以有C('z'-s[0], L-1)个,'z'-s[0] 是因为题意已经说明该字符串每个位置是严格递增的。如果在这个位置,还能找到别s[0]严格小的,则再加上。
再遍历到后面的字符s[i],对于比这个字符严格小的,我们依然可以找到C('z'-s[0], L-i-1)个。一直到最后一个字符,这样,所有结果的总和,就是结果。
对于1中的原理如下:
以两位的串为例,有ab~az,bc~cz,……那么有组合数
递推出这个公式,需要结合组合里的一个很常用的式子
后面的依此类推。有大佬也指出,既然你是升序,那么只要你有几位,你就选几个出来,那么它的严格升序排列只有一种,所以就是C(26,L),就是在长度为L下的所有组合数。
然后利用上面的式子进行递推,求出所有的26以内的所有组合数,再直接利用求结果即可。
AC代码:
#include <iostream> #include <cstdio> #include <cstring> using namespace std; typedef long long LL; LL C[27][27]; char S[15]; void getC() { memset(C, 0, sizeof(C)); for(int i = 0; i <= 26; i++) { for(int j = 0; j <= i; j++) { if(!j || i == j) C[i][j] = 1; else C[i][j] = C[i-1][j] + C[i-1][j-1]; } } C[0][0] = 1; } int main() { getC(); while(~scanf("%s", S)) { LL ans = 0; int len = strlen(S); //判断升序否 for(int i = 1; i < len; i++) { if(S[i] <= S[i-1]) { printf("0 "); return 0; } } //先把长度比S小的数目都统计加起来 for(int i = 1; i < len; i++) { ans += C[26][i]; } //统计长度相同的,枚举相加 for(int i = 0; i < len; i++) { int ch = i==0?'a':S[i-1]+1; //考虑到S升序 for(; ch < S[i]; ch++) { ans += C['z' - ch][len - i - 1]; //从能选的字母中选出len-i+1个进行组合 } } ans++; //加自己 printf("%lld ", ans); } return 0; }