POJ1850 组合数学

Code
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 8939 Accepted: 4264

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

题意：
字典序为a，b,c,…..ab，输入一个字符串，输出该字符串的位置
题解：
求长度为3满足题意得字符串总个数，假设第一个位置填一个字母，那么就需要再从25个字母里挑两个，也就是C（25，2），假设第一个位置换一个字母，那就再加上C（24，2），如此可知有C（25，2）+C（24，2）……C（2，2）= C（25，3），如此长度小于输入字符串的个数就确定了。接着，我们再求在相同长度时，该字符串排第几。一样的，前面的数确定了，我们挑几个数放进去就好。（因为挑出来的字母肯定只有一种严格递增的排列）

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <map>#include <queue>#include <vector>#define f(i,a,b) for(i = a;i<=b;i++)#define fi(i,a,b) for(i = a;i>=b;i--)using namespace std;int dp[40][40];char a[40];void init(){    int i,j,k;    memset(dp,0,sizeof(dp));    f(i,1,35) dp[i][0] = 1;    dp[1][1] = 1;    dp[0][0] = 0;    f(i,2,35)        f(j,1,i)            dp[i][j] = dp[i-1][j] + dp[i-1][j-1];}int main(){    init();    while(cin>>a){        int i,j,p = 1,cnt = 1;        for(i = 1;a[i]!='\0';i++){            if(a[i]<=a[i-1]){                p = 0;                printf("0\n");                return 0;            }            cnt++;        }        int ans = 0;        if(p){            f(i,1,cnt-1)                ans+=dp[26][i];            f(i,0,a[0]-'a'-1) ans+=dp[25-i][cnt-1];            f(i,1,cnt){                f(j,a[i-1]+1-'a',a[i]-1-'a'){                    ans += dp[25-j][cnt-(i+1)];                }            }        }        if(p == 0)            printf("0\n");        else            printf("%d\n",ans+1);    }    return 0;}