usaco/1.1 Your Ride Is Here

很纠结到底用C/C++还是java

对java更熟悉，C++除了基本语法，基本都还给老师了

但是不可否认，C++相对java效率更高，数据结构各种语言相差不大，就用C++吧

第一道题，题目很简单，遇到的困难竟然是输入输出流，习惯了java的输入输出流，c++的重新学习下

拿出《C++ primer》复习下标准IO库

IO类型在三个独立的头文件中定义:iostream定义读写控制窗口的类型，fstream定义读写已经命名文件的类型，而sstream所定义的类型则用于读写存储在内存中的string对象。

题目：

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.

Examples:

InputOutput

COMETQHVNGAT

GO

ABSTARUSACO 

STAY

提交文件

/*ID: chicc991PROG: rideLANG: C++*/#include <fstream>using namespace std;int main(){    ifstream fin("ride.in");    ofstream fout("ride.out");    char groupName[7];    char cometName[7];    int group = 1;    int comet = 1;    int i;    fin>>groupName>>cometName;    for(i=0; groupName[i]!='\0'; i++)    {        group*=groupName[i]-'A'+1;    }    for(i=0; cometName[i]!='\0'; i++)    {        comet*=cometName[i]-'A'+1;    }    if(group%47 == comet%47)    {        fout<<"GO"<<endl;    }    else fout<<"STAY"<<endl;    return 0;}

成功通过。

考虑到一些特殊情况，groupName[i]!='\0'改为i<a.size更好。