poj 1850 Code

Code Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 9366   Accepted: 4481 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 55 Source Romania OI 2002

这题我真是很不容易的自己做了 出来，然后看了看题解，就觉得自己的思路是正确的但在求解上他们用的杨辉三角更方便，哎！还是技不如人啊！

/*

大致题意：

输出某个str字符串在字典中的位置，由于字典是从a=1开始的，因此str的位置值就是 在str前面所有字符串的个数 +1

规定输入的字符串必须是升序排列。不降序列是非法字符串

不要求用循环输入去输入若干组字符串，但若输入非法字符串则输出0，且结束程序

*/
我主要是用了一个二维数组a[i][j]，i代表字符个数，j代表头字母的值,而具体的a[2][3]（cz的值）就代表字符为两个字符开头为C的最后一个的数值，

然后我就由此发现，

a[1][1] = 1, a[1][2] = 2, a[1][3] = 3``````

a[2][1] = a[1][26]+ a[1][26]-a[1][1],,,,,,,,a[2][2] = a[2][1]+ a[1][26]-a[1][2],,,,,,

a[3][1] = a[2][25]+ a[2][25]-a[2][1],,,   这里为什么是25呢？因为yz就是字符为2的最后一个了，

a[3][2] = a[3][1]+ a[2][25]-a[2][2],

以上我们可以在简化一下

令a[2][0] = a[1][26],

这样就更方便了如a[2][1] = a[2][0] + a[1][26]-a[1][1],

规律你发现了吗？

然后我们怎样计算bf的值呢，我们就用a[2][2]-(a[1][26]-a[1][5])就可以了，明白吗？

以下是代码：

#include <cstring>
#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;

int a[100][100];

void init()
{
int i, j;
memset(a, 0, sizeof(a));
for ( i = 1;i <= 26; i++ )
{
a[1][i] = i;
}
int h = 26;
for ( i = 2;i <= 10; i++ )
{
a[i][0] = a[i-1][h];
for ( j = 1; j <= h; j++ )
{
if ( j == 1 )
{
a[i][j] = a[i-1][h]+a[i-1][h]-a[i-1][j];
}
else
{
a[i][j] = a[i][j-1]+a[i-1][h]-a[i-1][j];
}
}
h--;
}
}

int Num(char ch[])
{
int i;
int len = strlen(ch);
int sum = a[len][ch[0]-'a'+1];
int sum2 = 0;
for ( i = 1;i < len; i++ )
{
sum2 = sum2+a[len-i][26-len+i]-a[len-i][ch[i]-'a'+1]+1;
}
return sum-sum2;
}

int main()
{
int i;
char ch[15];
init();
while (~scanf ( "%s", ch ) )
{
int len=strlen(ch);
for( i =1; i < len; i++ )
{
if( ch[i] <= ch[i-1] )
{
cout<<0<<endl;
return 0;
}
}
int sum = Num(ch);
printf ( "%d\n", sum );
}
}


代码菜鸟，如有错误，请多包涵！！！
如果有帮助记得支持我一下，谢谢！！！