Integer Factorization (30)解题报告

本文介绍了一个使用深度优先搜索(DFS)解决特定问题的C++实现案例。该算法应用于寻找一个整数可以被表示为几个较小整数的幂次之和的情况,并通过剪枝优化搜索过程以提高效率。

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利用DFS对状态空间进行搜索,注意进行剪枝,否则会超时。


#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <vector>
using namespace std;

int power(int num, int p);
void dfs(int num, int k, int p);

vector<int> v;
int upper;
int main(void) 
{
	int n, k, p, i;
	scanf("%d %d %d", &n, &k, &p);
	for (i = 1; i <= n; i++) {
		if (power(i, p) >= n) {
			upper = i;
			break;
		}
	}
	dfs(n, k, p);
	if (v.size()) {
		printf("%d = %d^%d", n, v[0], p);
		for (i = 1; i < v.size(); i++) {
			printf(" + %d^%d", v[i], p);
		}
		putchar('\n');
	}
	else {
		puts("Impossible");
	}
	return 0;
}

int power(int num, int p) 
{
	int i, sum = 1;
	for (i = 0; i < p; i++) {
		sum *= num;
	}
	return sum;
}

void dfs(int num, int k, int p)
{
	static vector<int> tmp;
	static int vsum = 0;
	if (!k && !num) {
		int sum = 0;
		if (!v.size()) {
			for (auto &i : tmp) {
				sum += i;
			}
			vsum = sum;
			v = tmp;
		}
		else {
			for (auto &i : tmp) {
				sum += i;
			}
			if (sum > vsum) {
				v = tmp;
				vsum = sum;
			}
			else if (sum == vsum && tmp > v) {
				v = tmp;
			}
		}
	}
	else if (num > 0 && k > 0) {
		int i, j;
		for (i = tmp.size() ? tmp[tmp.size() - 1] : upper; i >= 1; i--) {
			j = power(i, p);
			if (j * k < num) {
				break;
			}
			else if (j <= num) {
				tmp.push_back(i);
				dfs(num - j, k - 1, p);
				tmp.pop_back();
			}
		}
	}
	return;
}

{"id": "cmpl-e759f3d8-78e1-11f0-b2b4-1a85109d8f07", "choices": [{"finish_reason": "stop", "index": 0, "message": {"content": "<think>\nOkay, so I need to figure out how many whole-number divisors there are for the integer given by the expression $ a(2a + b) - 2a^2 + ab $, where $ a $ and $ b $ are different prime numbers greater than 2. Let me start by simplifying the expression step by step.\n\nFirst, let me write down the expression again to make sure I have it right: $ a(2a + b) - 2a^2 + ab $. Let me expand the first term, which is $ a(2a + b) $. If I distribute the $ a $, that becomes $ 2a^2 + ab $. So substituting that back into the original expression, we have:\n\n$ 2a^2 + ab - 2a^2 + ab $.\n\nNow let me combine like terms. The $ 2a^2 $ and $ -2a^2 $ cancel each other out, right? So that leaves me with $ ab + ab $, which is $ 2ab $. So the entire expression simplifies to $ 2ab $. \n\nWait, so the problem is asking for the number of whole-number divisors of $ 2ab $, where $ a $ and $ b $ are different primes greater than 2. Let me confirm that I simplified correctly. Let me check again:\n\nOriginal expression: $ a(2a + b) - 2a^2 + ab $.\n\nExpanding $ a(2a + b) $: $ 2a^2 + ab $. Then subtract $ 2a^2 $, so $ 2a^2 + ab - 2a^2 = ab $. Then add $ ab $, so $ ab + ab = 2ab $. Yes, that seems right. So the expression simplifies to $ 2ab $.\n\nAlright, so now I need to find the number of positive divisors of $ 2ab $. Since $ a $ and $ b $ are primes greater than 2, they are both odd primes. Let me recall that the number of divisors of a number is found by taking its prime factorization, adding one to each of the exponents, and multiplying those together.\n\nSo first, let me write the prime factorization of $ 2ab $. Since 2 is a prime number, and $ a $ and $ b $ are different primes, all three primes are distinct. Therefore, the prime factors of $ 2ab $ are 2, $ a $, and $ b $. Each of these primes has an exponent of 1 in the factorization. So the prime factorization is $ 2^1 \\times a^1 \\times b^1 $.\n\nTo find the number of divisors, I take each exponent (which are all 1s), add 1 to them, and multiply. That would be $ (1 + 1)(1 + 1)(1 + 1) = 2 \\times 2 \\times 2 = 8 $. Therefore, there are 8 positive divisors.\n\nBut let me verify this with an example to make sure I haven't made a mistake. Let me pick specific primes for $ a $ and $ b $. Let's take $ a = 3 $ and $ b = 5 $, which are both primes greater than 2 and different.\n\nCalculating $ 2ab $: $ 2 \\times 3 \\times 5 = 30 $. Now, the divisors of 30 are 1, 2, 3, 5, 6, 10, 15, 30. That's 8 divisors. Yep, that matches the calculation. Let me try another example to be safe. Let me take $ a = 5 $, $ b = 7 $. Then $ 2ab = 2 \\times 5 \\times 7 = 70 $. Divisors of 70: 1, 2, 5, 7, 10, 14, 35, 70. Again, 8 divisors. Okay, so that seems consistent.\n\nBut let me think again if there's any possibility that 2 could be equal to $ a $ or $ b $. However, the problem states that $ a $ and $ b $ are primes greater than 2, so they are odd primes. Therefore, 2 is not among them. Therefore, 2, $ a $, and $ b $ are distinct primes. Therefore, the prime factorization is indeed three distinct primes each to the first power, leading to 2×2×2=8 divisors.\n\nBut let me check if there's a case where maybe $ a $ or $ b $ is 2. But the problem says they are greater than 2, so they can't be. Therefore, my reasoning holds.\n\nTherefore, the number of whole-number divisors is 8. Therefore, the answer is 8, which in boxed form is \\boxed{8}.\n\nWait, but let me just check once more. Suppose someone might think that maybe the original expression simplifies differently? Let me check again:\n\nOriginal expression: $ a(2a + b) - 2a^2 + ab $.\n\nFirst term: $ a(2a + b) = 2a^2 + ab $.\n\nSubtracting $ 2a^2 $: $ 2a^2 + ab - 2a^2 = ab $.\n\nAdding $ ab $: $ ab + ab = 2ab $. Yes, that's correct. So the expression simplifies to 2ab. So no issues there.\n\nAlternatively, maybe someone might misread the original problem? Let me check again. The problem says: \"the integer $ a(2a + b) - 2a^2 + ab $\". So yes, that's how it's written. So the steps are correct.\n\nTherefore, I think my answer is correct. The number of divisors is 8.\n\n**Final Answer**\n\\boxed{8}\n</think>\n\nWe are given an expression involving two distinct prime numbers $ a $ and $ b $, both greater than 2, and asked to find the number of whole-number divisors of the integer:\n\n$$\na(2a + b) - 2a^2 + ab\n$$\n\n---\n\n### Step 1: Simplify the Expression\n\nStart by expanding the first term:\n\n$$\na(2a + b) = 2a^2 + ab\n$$\n\nNow substitute this back into the full expression:\n\n$$\n2a^2 + ab - 2a^2 + ab\n$$\n\nSimplify by combining like terms:\n\n- $ 2a^2 - 2a^2 = 0 $\n- $ ab + ab = 2ab $\n\nSo the entire expression simplifies to:\n\n$$\n2ab\n$$\n\n---\n\n### Step 2: Prime Factorization of $ 2ab $\n\nSince $ a $ and $ b $ are distinct prime numbers greater than 2, and 2 is also a prime, the expression $ 2ab $ is the product of three **distinct** primes:\n\n$$\n2ab = 2^1 \\cdot a^1 \\cdot b^1\n$$\n\n---\n\n### Step 3: Count the Number of Divisors\n\nThe number of positive divisors of a number is determined by taking the product of one more than each exponent in its prime factorization.\n\nHere, the exponents are all 1:\n\n$$\n(1 + 1)(1 + 1)(1 + 1) = 2 \\cdot 2 \\cdot 2 = 8\n$$\n\n---\n\n### Final Answer\n\n$$\n\\boxed{8}\n$$", "role": "assistant", "reasoning_content": ""}, "review": {"gold": "8", "pred": "8", "result": true}}], "created": 1755157209, "model": "Qwen3-32B", "object": "chat.completion", "usage": {"completion_tokens": 1627, "prompt_tokens": 68, "total_tokens": 1695}, "model_spec": {"api_url": "http://0.0.0.0:9009/v1/chat/completions", "model_id": "Qwen3-32B", "api_key": "EMPTY"}, "answer_id": "answer-1a07ae97260436f698e63d4620d7cff0", "subset_name": "default", "raw_input": {"problem": "Suppose $a$ and $b$ are different prime numbers greater than 2. How many whole-number divisors are there for the integer $a(2a+b)-2a^{2}+ab$?", "solution": "Distributing and combining like terms, we have $a(2a+b)-2a^2+ab=2a^2+ab-2a^2+ab=2ab$. Now $a$ and $b$ are different prime numbers greater than 2, so $2ab=2^1\\cdot a^1\\cdot b^1$ has $(1+1)(1+1)(1+1)=\\boxed{8}$ divisors.", "answer": "8", "subject": "Number Theory", "level": 4, "unique_id": "test/number_theory/1287.json"}, "origin_prompt": {"data": ["Suppose $a$ and $b$ are different prime numbers greater than 2. How many whole-number divisors are there for the integer $a(2a+b)-2a^{2}+ab$?\nPlease reason step by step, and put your final answer within \\boxed{}."], "system_prompt": null, "raw_input": {"problem": "Suppose $a$ and $b$ are different prime numbers greater than 2. How many whole-number divisors are there for the integer $a(2a+b)-2a^{2}+ab$?", "solution": "Distributing and combining like terms, we have $a(2a+b)-2a^2+ab=2a^2+ab-2a^2+ab=2ab$. Now $a$ and $b$ are different prime numbers greater than 2, so $2ab=2^1\\cdot a^1\\cdot b^1$ has $(1+1)(1+1)(1+1)=\\boxed{8}$ divisors.", "answer": "8", "subject": "Number Theory", "level": 4, "unique_id": "test/number_theory/1287.json"}, "index": 0}, "reviewed": true, "review_id": "review-307a3dc16716316d92e62e1f3ae09703", "reviewer_spec": {"metric": ["AveragePass@1"], "reviewer": ["Evaluator"], "revision": ["default"]}, "review_time": 1755194244.6042051} json文件更换为如上,分析并重新生成脚本,依旧生成csv文件
最新发布
08-16
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