USCAO-Section1.1 Your Ride Is Here

最新推荐文章于 2021-02-16 13:19:17 发布

原创最新推荐文章于 2021-02-16 13:19:17 发布 · 531 阅读

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#算法基础

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本文介绍了一个有趣的算法问题，通过将字符串转换为数值并进行模运算来判断两个字符串是否匹配。具体实现中，使用C++编程语言完成对输入字符串的处理，并依据特定规则输出“GO”或“STAY”。

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原题：
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.)
Write a program which reads in the name of the comet and the name of the group and figures out whether according to the above scheme the names are a match, printing “GO” if they match and “STAY” if not. The names of the groups and the comets will be a string of capital letters with no spaces or punctuation, up to 6 characters long.

题意: 输入两个字符串（1、纯大写字母 2、str.length()<=6）
字符[A-Z]->[1-26], 判断两个字符串的每个字符对应的数字累乘后模47的值是否一致，一致输出GO，不一致输出STAY，代码如下：

/*
ID:newyear111
PROG: ride
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <string>
#include<algorithm>
using namespace std;

int main()
{
    //文件读写  
    ofstream  fout ("ride.out");  
    ifstream  fin  ("ride.in");  
    string a,b;  
    while(fin>>a>>b)
    {
        //字符串最长6位  26^6不超出int的取值范围 
        int suma,sumb;
        suma=sumb=1;
        for(int i=0;i<a.length();i++)
            suma*=(a[i]-'A'+1);
        for(int i=0;i<b.length();i++)
            sumb*=(b[i]-'A'+1);
        suma%=47;sumb%=47;
        if(suma==sumb)
            fout<<"GO"<<endl;
        else
            fout<<"STAY"<<endl;
    }
    return 0;
}