C++ Problem P1200 [USACO1.1] Your Ride Is Here

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P1200 [USACO1.1] Your Ride Is Here

Your Ride Is Here

Problem Description

As is well known, there is a UFO behind every comet. These UFOs often come to collect loyal supporters from Earth. Unfortunately, their flying saucers can only take one group of supporters at a time. Therefore, they need a clever scheme to let these groups know in advance who will be taken away by the comet. They have given each comet a name, and based on these names, they decide whether a group is the specific one that will be taken away (who do you think named these comets?). The details of how the pairing works will be explained below; your task is to write a program that determines whether a group can be taken away by the UFO behind the comet based on the group name and the comet name.

Both the group name and the comet name are converted into a number in the following way: the final number is the product of all the letters in the name, where is , is . For example, the group is . If the group’s number $mod 47$ equals the comet’s number , you need to tell this group to be ready to be taken away! (Remember that “” is the remainder of divided by , for example, equals ).

Write a program that reads the comet name and the group name and calculates whether the two names can be paired according to the scheme above. If they can be paired, output GO; otherwise, output STAY. Both the group name and the comet name are strings of uppercase letters without spaces or punctuation (no more than letters).

Input Format

• Line 1: An uppercase letter string of length to representing the comet’s name.

• Line 2: An uppercase letter string of length to representing the group’s name.

Output Format

• None

Sample Input

#1

COMETQ

HVNGAT

#2

ABSTAR

USACO

Sample Output

#1

GO

#1

STAY

💡 Problem Analysis

This problem requires determining whether a group can be taken away by the UFO based on the special number conversion rules of the group name and comet name. The core is to convert the two names into numbers and compare their results modulo 47.

💡 Solution Approach

1. Input Reading: Read the comet name and group name strings.

2. Character to Number Conversion: Convert each letter to its corresponding number (A=1, B=2, …, Z=26).

3. Calculate Product Mod 47: Calculate the product of the letter numbers for each name and take modulo 47 (taking modulo stepwise to avoid overflow).

4. Compare Results: If the two modulo values are equal, output GO; otherwise, output STAY.

📐Mathematical Principles

• Letter to Number Mapping: Using ASCII characteristics, the number corresponding to letter c is c – ‘A’ + 1 (e.g., ‘A’-‘A’+1=1, ‘B’-‘A’+1=2).

• Properties of Modulo Operation: The modulo of a product equals the product of the modulos, i.e., (a*b) mod m = [(a mod m) * (b mod m)] mod m. This property allows taking modulo 47 stepwise during the product calculation to avoid large intermediate results leading to overflow.

🧰 Programming Principles

• String Traversal: Use a range for loop to traverse each character in the string.

• Variable Types: Use int to store the result of the product mod 47 (the maximum product of 6 letters is 26⁶=308915776, which is less than the maximum value of int 2147483647, so overflow is not a concern).

• Conditional Judgement: Use simple if-else to compare the two modulo values and output the corresponding result.

🧾Code Implementation

#include<iostream>   // Include input-output stream library for cin and cout#include<string>     // Include string processing library for string typeusing namespace std;  // Use standard namespace to simplify code writing
int main(){    string comet, group;  // Declare string variables to store comet name and group name    cin >> comet >> group; // Read two strings from standard input
    int cometMod = 1;      // Initialize the product mod 47 result for comet name (initial value for multiplication is 1)    for (char c : comet) { // Traverse each character c in the comet name string        int num = c - 'A' + 1; // Convert character to number: e.g., 'C'-'A'+1=3        cometMod = (cometMod * num) % 47; // Multiply and take mod 47 to prevent overflow    }
    int groupMod = 1;      // Initialize the product mod 47 result for group name    for (char c : group) { // Traverse each character c in the group name string        int num = c - 'A' + 1; // Convert character to number        groupMod = (groupMod * num) % 47; // Multiply and take mod 47    }
    if (cometMod == groupMod) { // Compare whether the two modulo values are equal        cout << "GO" << endl;   // If equal, output "GO"    } else {        cout << "STAY" << endl; // If not equal, output "STAY"    }
    return 0; // Program ends normally}

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Direct access to the Luogu problem bank for practice

🛠️ Programming Tips

• Range for loop: for (char c : str) simplifies traversing the string, avoiding manual index control.

• Character to Number Conversion: Quickly calculate the corresponding number for letters using c – ‘A’ + 1 without complex conditional checks.

• Stepwise Modulo: Immediately take modulo 47 after each multiplication, which conforms to mathematical principles and avoids overflow of intermediate results.

📚 Function / Method Usage Instructions

📚 Vocabulary Cards for Variables and Functions

⏱️ Time & Space Complexity Analysis

• Time Complexity: O(n + m), where n and m are the lengths of the comet name and group name, respectively. Each character needs to be traversed once, and the operations (character conversion, multiplication, modulo) in each traversal are O(1).

• Space Complexity: O(1), only a fixed number of variables are used (the space for storing strings is independent of input length, considered constant).

🧩 Problem Summary

This problem tests string processing, properties of modulo operations, and basic logical judgment abilities. The core is to convert a textual problem into mathematical calculations, utilizing the properties of modulo operations to optimize the calculation process. By traversing strings, converting characters to numbers, and taking modulo stepwise, the problem can be solved efficiently. The key is to understand the conversion rules of “product mod 47” and the simplification effect of modulo operations.

END

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