usaco/1.1 Your Ride Is Here

很糾結到底用C/C++還是java

對java更熟悉，C++除了基本語法，基本都還給老師了

但是不可否認，C++相對java效率更高，數據結構各種語言相差不大，就用C++吧

第一道題，題目很簡單，遇到的困難竟然是輸入輸出流，習慣了java的輸入輸出流，c++的重新學習下

拿出《C++ primer》複習下標準IO庫

IO類型在三個獨立的頭文件中定義:iostream定義讀寫控制窗口的類型，fstream定義讀寫已經命名文件的類型，而sstream所定義的類型則用於讀寫存儲在內存中的string對象。

題目：

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:

Input	Output
COMETQ HVNGAT	GO
ABSTAR USACO	STAY

提交文件

/*
ID: chicc991
PROG: ride
LANG: 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;
}

成功通過。

考慮到一些特殊情況，groupName[i]!='\0'改爲i<a.size更好。