求整数n的最小质因子出现次数 打表

Posted 2022-03-18 karshey

去年校赛没做出来，今年能做出来了。补个题解。

可以直接欧拉筛，找到输入数字的第一个质因数。
3e6以内的质因数不会大于sqrt(se6)=1733，也就是最多会有1733个质数，实际上只会有269个质数。
欧拉筛，外层是sqrt(se6)=1733，内层是3e6，一共3e9，直接筛可能会TLE（不过实际上提交貌似不会），保险起见我们把质数打表打出来。

打表函数：

#include<bits/stdc++.h>
using namespace std;
const int N=3e6+10;
map<int,int>v;
int p[300];
int n,idx;
void ini()

	for(int i=2;i*i<=3000000;i++)
	
		if(!v[i])
		
			p[idx++]=i;			
			for(int j=i*i;j<=3000000;j+=i)
			
				v[j]=1;
			
		
	

int main()

	ini();
	for(int i=0;i<idx;i++)
	
		cout<<p[i]<<",";
	
	return 0;

打出来的为：这就是3e6以内所有合数的质因数了，如果不是，则说明这个数本身就是质数，输出1.

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

提交的代码：

#include<bits/stdc++.h>
using namespace std;
int n;
int p[300]=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;
//0-268
int main()

	scanf("%d",&n);
	while(n--)
	
		int t;scanf("%d",&t);
		if(t==1) 
		
			printf("0\\n");continue;
		
		int tt=-1;
		for(int i=0;i<269;i++)
		
			if(t%p[i]==0)
			
				tt=p[i];break;
			
		
		if(tt==-1)
		
			printf("1\\n");//它自己是质数
			continue; 
		
		int ans=0;
		while(t%tt==0)
		
			t/=tt;ans++;
		
		printf("%d\\n",ans);
	
	return 0;