UVa 417-Word Index_word417-优快云博客

本文链接：https://blog.youkuaiyun.com/AcToy/article/details/8981509

本文介绍了一个通过字母组合生成固定数值的程序，利用打表方式实现不同长度字符串到数值的快速转换。

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根据所给的方式每种字母的组合都对应一个固定的数值，所以我采用的打表的方式，没仔细想其他的方法。

/*************************************************************************
    > File Name: 417.cpp
    > Author: Toy
    > Mail: ycsgldy@163.com 
    > Created Time: 2013年05月27日 星期一 19时18分53秒
 ************************************************************************/

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <cstdlib>
#include <climits>
#include <sstream>
#include <fstream>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>

using namespace std;
const int INF = 0x7fffffff;
char str[7];
int a2[30][30], a3[30][30][30], a4[30][30][30][30], a5[30][30][30][30][30];

int main ( ) {
	int cnt = 27;
	for ( int i = 1; i <= 26; ++i ) 
		for ( int j = i + 1; j <= 26; ++j ) 
			a2[i][j] = cnt++;
	for ( int i = 1; i <= 26; ++i ) 
		for ( int j = i + 1; j <= 26; ++j ) 
			for ( int k = j + 1; k <= 26; ++k ) 
				a3[i][j][k] = cnt++;
	for ( int i = 1; i <= 26; ++i ) 
		for ( int j = i + 1; j <= 26; ++j ) 
			for ( int k = j + 1; k <= 26; ++k ) 
				for ( int m = k + 1; m <= 26; ++m ) 
					a4[i][j][k][m] = cnt++;
	for ( int i = 1; i <= 26; ++i ) 
		for ( int j = i + 1; j <= 26; ++j ) 
			for ( int k = j + 1; k <= 26; ++k ) 
				for ( int m = k + 1; m <= 26; ++m ) 
					for ( int n = m + 1; n <= 26; ++n ) 
						a5[i][j][k][m][n] = cnt++;
	while ( scanf ( "%s", str ) == 1 ) {
		int len = strlen ( str );
		bool flag = 1;
		for ( int i = 1; i < len; ++i ) 
			if ( str[i] < str[i - 1] ) flag = 0;
		if ( !flag ) printf ( "0\n" );
	    else {
			if ( len == 1 ) printf ( "%d\n", str[0] - 'a' + 1 );
			else if ( len == 2 ) printf ( "%d\n", a2[str[0] - 'a' + 1][str[1] - 'a' + 1] );
			else if ( len == 3 ) printf ( "%d\n", a3[str[0] - 'a' + 1][str[1] - 'a' + 1][str[2] - 'a' + 1] );
			else if ( len == 4 ) printf ( "%d\n", a4[str[0] - 'a' + 1][str[1] - 'a' + 1][str[2] - 'a' + 1][str[3] - 'a' + 1] );
			else if ( len == 5 ) printf ( "%d\n", a5[str[0] - 'a' + 1][str[1] - 'a' + 1][str[2] - 'a' + 1][str[3] - 'a' + 1][str[4] - 'a' + 1] );
		}	
	}
    return 0;
}