UVa 10105 Polynomial Coefficients(组合数学)

关注

原创最新推荐文章于 2026-03-06 00:16:43 发布 1.3k 阅读 AI 写同款· GEO 优化 ›

快手2019校招工程B卷复盘：大厂笔试核心考点与备考策略在互联网大厂技术校招中，笔试环节往往综合考察候选人的算法功底、计算机基础与工程实践能力。从数据结构中的HashMap与动态规划，到计算机网络里的TCP状态变迁，再到数据库索引与事务隔离级别，这些高频考点本质上都在检验开发者是否具备系统性的技术认知。系统设计题则进一步要求候选人将基础原理转化为可落地的架构方案，如短链接服务或推送系统的设计。通过拆解快手2019年秋季校园招聘工程B试卷，可以清晰看到大厂笔试的出题逻辑与筛选标准——不仅是知识记忆，更是对问题建模、边界处理和技术权衡的全面评估。本文从试卷结构、核心阅读详情

10105 - Polynomial Coefficients

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=115&page=show_problem&problem=1046

The Problem

The problem is to calculate the coefficients in expansion of polynomial(x₁+x₂+...+x_k)ⁿ.

The Input

The input will consist of a set of pairs of lines. The first line of the pair consists of two integers n and k separated with space (0<K,N<13). This integers define the power of the polynomial and the amount of the variables. The second line in each pair consists of k non-negative integers n₁, ..., n_k, where n₁+...+n_k=n.

The Output

For each input pair of lines the output line should consist one integer, the coefficient by the monomial x₁ⁿ¹x₂ⁿ²...x_k^nk in expansion of the polynomial(x₁+x₂+...+x_k)ⁿ.

Sample Input

2 2
1 1
2 12
1 0 0 0 0 0 0 0 0 0 1 0

Sample Output

2
2

题意&思路：参考Richard A.Brualdi的《组合数学（第5版）》P88 (5.4 多项式定理)

完整代码：

非打表：

/*0.016s*/

#include <cstdio>
using namespace std;

int main()
{
	int n, k, ni;
	while (~scanf("%d%d", &n, &k))
	{
		long long sum = 1;
		for (int i = 2; i <= n; i++)
			sum *= i;
		while (k--)
		{
			scanf("%d", &ni);
			for (int i = 2; i <= ni; i++)
				sum /= i;
		}
		printf("%lld\n", sum);
	}
	return 0;
}

打表：

/*0.015s*/

#include <cstdio>
using namespace std;
const long long f[11] = {6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, 6227020800};

int main()
{
	int n, k, ni;
	while (~scanf("%d%d", &n, &k))
	{
		long long sum = 1;
		if (n == 2)
			sum <<= 1;
		else if (n > 2)
			sum *= f[n - 3];
		while (k--)
		{
			scanf("%d", &ni);
			if (ni == 2)
				sum >>= 1;
			else if (ni > 2)
				sum /= f[ni - 3];
		}
		printf("%lld\n", sum);
	}
	return 0;
}