『NOIP2016』T4 problem

最新推荐文章于 2024-07-18 22:34:42 发布

Z_Dex

最新推荐文章于 2024-07-18 22:34:42 发布

阅读量506

点赞数

CC 4.0 BY-SA版权

分类专栏：数学(快速幂/矩阵乘/素数/Gcd) 竞赛历程

本文链接：https://blog.youkuaiyun.com/Z_Dex/article/details/53400199

竞赛历程同时被 2 个专栏收录

2 篇文章

订阅专栏

数学(快速幂/矩阵乘/素数/Gcd)

1 篇文章

订阅专栏

本文介绍了一种利用质因数分解求解组合数的算法，通过杨辉三角和质因数加减实现组合数计算，并判断能否被特定数整除。最终通过矩阵前缀和得出答案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

『题解』

组合数算法有很多种，可以是杨辉三角，也可以是分解质因数。这题我开始就想到分解质因数，没想到杨辉三角。对于N！质因数都是（N-1)! 的质因数+ N的质因数。

在用公式求组合数就可以用加减质因数。判断每个组合数是否能整除K，就用其质因数减K的质因数，若出现小于0 的，即不能。最后用矩阵前缀和求出答案即可。

#include
#include
#include
#include
#include
#include

using namespace std;

queue q;

struct Tnum
{
	int prime[2001];
	int t, num, Min;
	Tnum()
	{
		Min = INT_MAX;
	}
}p[2010], M;

int prime[] = {0, 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997,1009,1013,1019,1021,1031,1033,1039,1049,1051,1061,1063,1069,1087,1091,1093,1097,1103,1109,1117,1123,1129,1151,1153,1163,1171,1181,1187,1193,1201,1213,1217,1223,1229,1231,1237,1249,1259,1277,1279,1283,1289,1291,1297,1301,1303,1307,1319,1321,1327,1361,1367,1373,1381,1399,1409,1423,1427,1429,1433,1439,1447,1451,1453,1459,1471,1481,1483,1487,1489,1493,1499,1511,1523,1531,1543,1549,1553,1559,1567,1571,1579,1583,1597,1601,1607,1609,1613,1619,1621,1627,1637,1657,1663,1667,1669,1693,1697,1699,1709,1721,1723,1733,1741,1747,1753,1759,1777,1783,1787,1789,1801,1811,1823,1831,1847,1861,1867,1871,1873,1877,1879,1889,1901,1907,1913,1931,1933,1949,1951,1973,1979,1987,1993,1997,1999};
int t, k, tp[10010][2], MaxN, MaxM, ans[2010][2010], Ans[2010][2010];

int work(int x)
{
	int Tp = 1, Pr;
	for (int i = 1;i <= x; i++)
	{
		for (int j = 1;prime[j] <= x; j++)
		{
			p[i].prime[j] = p[i - 1].prime[j];
		}
		Tp = 1;
		Pr = i;
		while (prime[Tp] <= i)
		{
			while (Pr % prime[Tp] == 0)
			{
				p[i].prime[Tp] ++;
				Pr /= prime[Tp];
				p[i].t = max(p[i].t, Tp);
			}
			Tp++;
		}
	}
	return 0;
}

int M_prime()
{
	int Tp = 1, Pr = M.num;
	while (prime[Tp] <= M.num)
	{
		while (Pr % prime[Tp] == 0)
		{
			Pr /= prime[Tp];
			M.prime[Tp]++;
			M.t = max(M.t, Tp);
		}
		Tp++;
	}
	return 0;
}

int Find(int x ,int mid)
{
	for (int i = 1;i <= M.t; i++)
	{
		if (p[x].prime[i] - p[mid].prime[i] - p[x - mid].prime[i] - M.prime[i] < 0) return 0;
	}
	return 1;
}

int main(void)
{
	freopen("problem.in", "r", stdin);
	freopen("problem.out", "w", stdout);
	
	cin >> t >> M.num;
	for (int i = 1;i <= t; i++)
	{
		scanf("%d%d", &tp[i][1], &tp[i][2]);
		MaxN = max(MaxN, tp[i][1]);
		MaxM = max(MaxM, tp[i][2]);
	}
	MaxM = min(MaxN, MaxM);
	work(MaxN);
	M_prime();
	for (int i = 1;i <= MaxN; i++)
	{
		for (int j = 1;j <= i; j++)
			ans[i][j] = Find(i, j);
	}
	for (int i = 1;i <= MaxN; i++)
	{
		for (int j = 1;j <= MaxN; j++)
		{
			Ans[i][j] = Ans[i - 1][j] + Ans[i][j - 1] - Ans[i - 1][j - 1] + ans[i][j];
		}
	}
	for (int i = 1;i <= t; i++)
		printf("%d\n", Ans[tp[i][1]][min(tp[i][1], tp[i][2])]);
	
	fclose(stdin);
	fclose(stdout);
	return 0;
}

By：Z_Dex