Your Ride Is Here

最新推荐文章于 2022-12-12 19:48:32 发布

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#usaco解题报告

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本文介绍了一道编程题，题目要求将特定的字符串转换成数值，并比较两个字符串转换后的数值是否相同模47。提供了Python3的示例代码，展示了如何读取输入文件、处理字符串并写出结果。

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题目描述：

It is a well-known fact that behind every good comet is a UFO. These UFOs often come to collect loyal supporters from here on Earth. Unfortunately, they only have room to pick up one group of followers on each trip. They do, however, let the groups know ahead of time which will be picked up for each comet by a clever scheme: they pick a name for the comet which, along with the name of the group, can be used to determine if it is a particular group's turn to go (who do you think names the comets?). The details of the matching scheme are given below; your job is to write a program which takes the names of a group and a comet and then determines whether the group should go with the UFO behind that comet.

Both the name of the group and the name of the comet are converted into a number in the following manner: the final number is just the product of all the letters in the name, where "A" is 1 and "Z" is 26. For instance, the group "USACO" would be 21 * 19 * 1 * 3 * 15 = 17955. If the group's number mod 47 is the same as the comet's number mod 47, then you need to tell the group to get ready! (Remember that "a mod b" is the remainder left over after dividing a by b; 34 mod 10 is 4.)

输入输出格式：

SAMPLE INPUT (file ride.in)

COMETQ
HVNGAT

OUTPUT FORMAT

A single line containing either the word "GO" or the word "STAY".

SAMPLE OUTPUT (file ride.out)

GO

题目大意

将字符串转变为数字，字母A对应的值为1，依次对应，字母Z对应的值为26。现在有一个字符串，将其中的每个字符转变为数字之后进行累乘，最终的结果对47求余数。
题目给你两个字符串，其中的字符都是大写字母，如果通过上述处理，两个字符串最终转变的数字结果是相等的，输出GO，否则输出STAY。

解题思路：

直接用对应的字母相减，然后做一个取余运算就行

解题代码：

我是使用python3来解决usaco的问题

fin = open ('ride.in', 'r')
fout = open ('ride.out', 'w')
x=fin.readline().strip('\n')
y=fin.readline().strip('\n')
sumx=1
sumy=1
for i in range(len(x)):
    sumx*=(ord(x[i])-(ord('A')-1))
for i in  range(len(y)):
    sumy*=(ord(y[i])-(ord('A')-1))
outx=sumx%47;
outy=sumy%47;
if outx==outy:
    fout.write("GO\n")
else:
    fout.write("STAY\n")
fout.close()