POJ 题目1850 Code（组合数学）_transmitting and memorizing information is a task…-优快云博客

本文链接：https://blog.youkuaiyun.com/yu_ch_sh/article/details/42612773

本文介绍了一种针对递增字符串的编码算法，并提供了一个AC代码示例。该算法首先检查输入字符串是否为递增字符串，然后计算所有比该字符串小或相等的递增字符串的数量作为其编码。

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Code

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 8404		Accepted: 3976

Description

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.

Input

The only line contains a word. There are some constraints:
• 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

Source

Romania OI 2002

题目大意是舒输入个递增的字符串，问小于等于他的递增字符串的个数。如果不是合法的就输出0就好

思路详见：http://blog.youkuaiyun.com/lyy289065406/article/details/6648492

ac代码

#include<stdio.h>
#include<string.h>
#include<math.h>
int c[30][30];
void fun()
{
	int i,j;
	for(i=0;i<30;i++)
	{
		c[i][0]=1;
	}
	for(i=1;i<30;i++)
	{
		for(j=1;j<30;j++)
			c[i][j]=c[i-1][j-1]+c[i-1][j];
	}
}
int main()
{
	char s[20];
	fun();
	while(scanf("%s",s)!=EOF)
	{
		int len=strlen(s);
		__int64 sum=0;
		int i,j,k;
		for(i=0;i<len-1;i++)
		{
			if(s[i]>s[i+1])
			{
				printf("0\n");
				return 0;
			}
		}
		for(i=1;i<len;i++)
		{
			sum+=c[26][i];
		}
		for(i=0;i<len;i++)
		{
			int ch;
			if(i==0)
				ch='a';
			else
				ch=s[i-1]+1;
			while(ch<s[i])
			{
				sum+=c['z'-ch][len-i-1];
				ch++;
			}
		}
		printf("%I64d\n",++sum);

	}
}