标签：acm   c   uva   源代码   算法   

说说：这道题非常之坑！开始还以为要递归判断的，其实根本不用，只要判断第一层的S即可，其中内嵌的S肯定是正确的。居然还有这样的题目，真的很无语╮(╯_╰)╭

题目：

Secret Research

At a certain laboratory results of secret research are thoroughly encrypted. A result of a single experiment is stored as an information of

实验室关于一些秘密研究成果都是进行完全加密的。实验的结果按照它的完成情况分为：

 its completion:

`positive result‘, `negative result‘, `experiment failed‘ or `experiment not completed‘

‘积极结果’，‘消极结果’，‘实验失败’，‘实验未完成’

The encrypted result constitutes a string of digits S, which may take one of the following forms:

加密的结果组成了一个数字串S，它将采用以下的某种形式

 positive result 		 S = 1 or S = 4 or S = 78  negative result 		 S = S35  experiment failed 		 S = 9S4  experiment not completed 		 S = 190S 

(A sample result S35 means that if we add digits 35 from the right hand side to a digit sequence then we shall get the digit sequence corresponding to a failed experiment)

(如结果S35意思是如果我们把数字35加到一个数字串的右边那么这就是个消极结果（译：原文为实验失败，我认为有误）)

You are to write a program which decrypts given sequences of digits.

你需要写一个程序对数字串进行解密

Input 

整数n代表加密结果的数目，接下来的连续的n行，每行都是用ASCII码表示的数字串

Output 

分析完数字串后需要输出结果，且一行一个结果

+ for a positive result - for a negative result * for a failed experiment ? for a not completed experiment 

In case the analysed string does not determine the experiment result, a first match from the above list should be outputted.

万一数字串并为决定实验结果，输出上表中第一个匹配的项。

Sample Input 

4
78
7835
19078
944

Sample Output 

+
-
?
*

源代码：

#include <stdio.h>
#include <string.h>

int main(){
 int N,len;
 char S[1000];
 //freopen("data.txt","r",stdin);
 scanf("%d",&N);

 while(N--){
    scanf("%s",S);
    len=strlen(S);
    if(len<=2){//是否为positive
        printf("+\n");
        continue;
    }
    if(S[0]=='9'&&S[len-1]=='4'){//是否为failed
        printf("*\n");
        continue;
    }
    if(S[len-1]=='5'){//是否为negative
        printf("-\n");
        continue;
    }
    printf("?\n");
 }
 return 0;
}

Secret Research UVA 621,布布扣,bubuko.com

Secret Research UVA 621

标签：acm   c   uva   源代码   算法   

原文地址：http://blog.csdn.net/u011915301/article/details/38406469