POJ 1580 Code

Code

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 6453 Accepted: 3026

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

这题我一开始因为一个字符写错了，不断的wa，我都无奈了，最后才注意到一个字符的错误，这个题最好用杨辉三角求组合数，否者要考虑精度问题。 注释掉的是我用的要考虑精度的算法。 这种算法相对来说比较耗时，每次要求组合数的时候都要重新进行，不如 杨辉三角这接打表 省时

#include <stdio.h>#include <string.h>#include <math.h>char s1[20];int zuhe[27][27];int main(){    long long int f(int m,int n);    void get_zuhe();    int i,j,n,m,t,l;    long long int s;    get_zuhe();    while(scanf("%s",s1)!=EOF)    {        l=strlen(s1);        for(i=1; i<=l-1; i++)        {            if(s1[i]<=s1[i-1])            {                printf("0\n");               break;            }        }        if(i!=l)        {            continue;        }        if(l==1)        {            printf("%d\n",s1[0]-'a'+1);        }        else        {            s=0;            for(i=1; i<=l-1; i++)            {                s+=f(i,26);            }            for(i=0; i<=l-1; i++)            {                if(i==0)                {                    t=s1[i]-'a';                }                else                {                    t=s1[i]-s1[i-1]-1;                }                if(i!=l-1)                {                    for(j=1; j<=t; j++)                    {                        if(i==0)                        {                            s+=f(l-1-i,26-j);                        }                        else                        {                            s+=f(l-1-i,'z'-(s1[i-1]+j));                        }                    }                }            }            s+=(s1[l-1]-s1[l-2]);            printf("%lld\n",s);        }    }    return 0;}void get_zuhe(){    int i,j;    memset(zuhe,0,sizeof(zuhe));    for(i=0; i<=26; i++)    {        for(j=0; j<=i; j++)        {            if(!j||i==j)zuhe[i][j]=1;            else zuhe[i][j]=zuhe[i-1][j-1]+zuhe[i-1][j];        }    }}long long int f(int m,int n){    long long int s;    s=zuhe[n][m];    return s;}/*long long int f(int m,int n){    int i,j;    long long int res;    double s=1.0;    for(i=n,j=m;i>=n-m+1;i--,j--)    {        s=s*(double)i/(double)j;    }    res=(long long int)(s+0.01);    return res;} */