组合数学....和上一题是一样的.... Word Index Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 4303 Accepted: 2439 Description Encoding schemes are often used in situations requiring encryption or information storage/transmission economy. Here, we develo…

Code Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 8256 Accepted: 3906 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 t…

Description   1942 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 considere…

http://poj.org/problem?id=1850 -----------------http://poj.org/problem?id=1496 两题解法类似..本题为组合数学的题,要求所给字符串在排列中的位置.所用的方法为暴力枚举. 对于长度为1的字符串有a,b,c,d.......26个:长度为2的字符串有 a开头时有ab,ac,ad,ae,af.....(25,1) 25个 //()为组合数 b开头时有bc,bd,de,.....(24,1) 24个 c开头时有cd,ce,cf…

Word Puzzles Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 10782 Accepted: 4076 Special Judge Description Word puzzles are usually simple and very entertaining for all ages. They are so entertaining that Pizza-Hut company started using t…

题目链接:http://poj.org/problem?id=1204 思路分析:由于题目数据较弱,使用暴力搜索:对于所有查找的单词建立一棵字典树,在图中的每个坐标,往8个方向搜索查找即可: 需要注意的是查找时不能匹配了一个单词就不在继续往该方向查找,因为在某个坐标的某个方向上可能会匹配多个单词,所以需要一直 查找直到查找到该方向上最后一个坐标: 代码如下: #include <cstdio> #include <cstring> #include <iostream>…

Description Word puzzles are usually simple and very entertaining for all ages. They are so entertaining that Pizza-Hut company started using table covers with word puzzles printed on them, possibly with the intent to minimise their client's percepti…

Word Amalgamation Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9968   Accepted: 4774 Description In millions of newspapers across the United States there is a word game called Jumble. The object of this game is to solve a riddle, but…

#include <iostream> #include <string> using namespace std; int fac(int num); int C(int n,int m); int com( int n, int k ); int main() { //freopen("acm.acm","r",stdin); string s; int sum; int j; int i; // cout<<com(4,2)…

poj1850 Code 题意:输出若干个给定的字符串($length<=10$)在字典序中的位置,字符串中的字母必须严格递增. 读取到非法字符串时,输出“0”,终止程序.(poj1496:继续读取) 我们分成2种情况讨论字典序小于给定字符串的字符串个数 1.长度比给定字符串小 其实长度为$i$的字符串的个数就是$C(26,i)$ 因为我们随机选取$i$个字符后,从小到大依次取出,是可以保证字符串是有序的. 2.长度等于给定字符串 我们枚举每一位,计算从该位开始小于给定字符串的个数,同样可以用组…