UVA621 Secret Research【水题】
作者:互联网
At a certain laboratory results of secret research are thoroughly encrypted. A result of a single experiment is stored as an information of its completion:
‘positive result’, ‘negative result’, ‘experiment failed’ or ‘experiment not completed’
The encrypted result constitutes a string of digits S, which may take one of the following forms:
• 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)
You are to write a program which decrypts given sequences of digits.
Input
A integer n stating the number of encrypted results and then consecutive n lines, each containing a sequence of digits given as ASCII strings.
Output
For each analysed sequence of digits the following lines should be sent to output (in separate lines):
+ 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.
Sample Input
4
78
7835
19078
944
Sample Output
- -
?
问题链接:UVA621 Secret Research
问题简述:(略)
问题分析:
给定一个字符串,若为“1”,“4”,“78”则输出‘+’;若最后两个字符为“35”则输出‘-’;若首字符为‘9’并且尾字符为‘4’则输出‘*‘;否则输出’?‘。
简单的字符串判定题,用C语言的字符串库函数(string.h)就可以实现。
程序说明:(略)
参考链接:(略)
题记:(略)
AC的C++语言程序如下:
/* UVA621 Secret Research */
#include <bits/stdc++.h>
using namespace std;
const int N = 512;
char s[N];
int main()
{
int n, len;
scanf("%d", &n);
while(n--) {
scanf("%s", s);
len = strlen(s);
if(strcmp(s, "1") == 0 || strcmp(s, "4") == 0 || strcmp(s, "78") == 0)
putchar('+');
else if(s[len - 2] == '3' && s[len - 1] == '5')
putchar('-');
else if(s[0] == '9' && s[len - 1] == '4')
putchar('*');
else
putchar('?');
putchar('\n');
}
return 0;
}
标签:digits,putchar,水题,len,failed,Secret,experiment,result,UVA621 来源: https://www.cnblogs.com/tigerisland45/p/10391911.html