中文繁簡體轉換,請在ChatGPT Custom instructions
1.What would you like ChatGPT to know about you to provide better responses?(描述自己的背景或情境)
2.How would you like ChatGPT to respond?(調整 ChatGPT 的輸出方式,例如語言、細節)
在2.加入語言=繁體中文
ChatGPT 問題輸入:
大學程式能力檢定 UVA100 The 3n + 1 problem 問題的中文與英文描述,包含錯誤處理和邊界檢查的完整C++範例程式碼,用於解決UVA100問題,以及解題思路的解釋
題目描述:
中文:對於給定的一個正整數n,我們定義一個遞迴數列如下:
這個遞迴過程持續進行,直到n變為1為止。例如,對於n = 22,這個遞迴過程如下:
22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
我們想知道對於給定範圍內的所有正整數,哪個數的遞迴過程生成的數字最多。
英文:Consider the following algorithm to generate a sequence of numbers. Start with an integer n:
Repeat the process with the new value of n, terminating when n becomes 1. For example, the following sequence of numbers will be generated for n = 22:
22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
Given an input n, it is possible to determine the number of numbers generated before 1 is generated. For example, for n = 22, the sequence length is 16.
You are tasked with finding the integer within a certain range that generates the longest sequence before reaching 1 and outputting both the starting number and the length of the sequence.
範例程式碼:
#include <iostream>
#include <map>
using namespace std;
int getSequenceLength(long long n, map<long long, int>& cache) {
if (n == 1) {
return 1;
}
if (cache.find(n) != cache.end()) {
return cache[n];
}
int length;
if (n % 2 == 0) {
length = 1 + getSequenceLength(n / 2, cache);
} else {
length = 1 + getSequenceLength(3 * n + 1, cache);
}
cache[n] = length;
return length;
}
int main() {
long long n;
while (cin >> n) {
long long start = min(n, 1LL); // Initialize the start and end values
long long end = max(n, 1LL);
int maxLength = 0;
map<long long, int> cache;
for (long long i = start; i <= end; i++) {
int length = getSequenceLength(i, cache);
maxLength = max(maxLength, length);
}
cout << n << " " << maxLength << endl;
}
return 0;
}
這個程式碼會計算在給定範圍內的所有正整數中,哪個數的遞迴過程生成的數字最多,並輸出該數字以及生成的數字的最大長度。
解題思路:
ChatGPT 問題輸入:
大學程式能力檢定 UVA10929 You can say 11 問題的中文與英文描述,包含錯誤處理和邊界檢查的完整C++範例程式碼,用於解決UVA10929問題,以及解題思路的解釋
題目描述:
中文:給定一個正整數n,你的任務是判斷n是否可以被11整除,如果可以,輸出"是",否則輸出"不是"。
英文:Given a positive integer n, your task is to determine if n is divisible by 11. If it is divisible, output "YES," otherwise, output "NO."
範例程式碼:
#include <iostream>
#include <string>
using namespace std;
bool isDivisibleBy11(const string& number) {
int oddSum = 0;
int evenSum = 0;
bool isOdd = true;
for (char digit : number) {
if (digit != ' ') {
int digitValue = digit - '0';
if (isOdd) {
oddSum += digitValue;
} else {
evenSum += digitValue;
}
isOdd = !isOdd;
}
}
int difference = abs(oddSum - evenSum);
return difference % 11 == 0;
}
int main() {
string number;
while (cin >> number) {
if (number == "0") {
break; // Input ends with "0"
}
bool divisibleBy11 = isDivisibleBy11(number);
if (divisibleBy11) {
cout << number << " is a multiple of 11." << endl;
} else {
cout << number << " is not a multiple of 11." << endl;
}
}
return 0;
}
這個程式碼會讀取一個正整數n,然後檢查它是否可以被11整除,並輸出相應的結果。
解題思路:
isDivisibleBy11
,它接受一個表示正整數的字串作為輸入。isDivisibleBy11
函數判斷每個輸入的正整數是否可以被11整除,並輸出相應的結果。