质数

质数距离

l,r,都非常大，但$l-r$很小所以晒出$1至sqrt(r)$的质数，再用倍数法求解
注意：质数为$i$时在数组中村$i-l$,不然数组会爆

#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
using namespace std;
const int N = 100006, L = 1000006, M = 46340, INF = 0x7fffffff;
bool v[L];
vector<int> p, ans;

int main() {
	memset(v, 1, sizeof(v));
	for (int i = 2; i <= M; i++)
		if (v[i]) {
			p.push_back(i);
			for (int j = 2; j <= M / i; j++) v[i*j] = 0;
		}
	unsigned int l, r;
	while (cin >> l >> r) {
		memset(v, 1, sizeof(v));
		ans.clear();
		if (l == 1) v[0] = 0;
		for (unsigned int i = 0; i < p.size(); i++)
			for (unsigned int j = (l - 1) / p[i] + 1; j <= r / p[i]; j++)
				if (j > 1) v[p[i]*j-l] = 0;
		for (unsigned int i = l; i <= r; i++) 
			if (v[i-l]) ans.push_back(i);
		int minn = INF, maxx = 0, x1, y1, x2, y2;
		for (unsigned int i = 0; i + 1 < ans.size(); i++) {
			int num = ans[i+1] - ans[i];
			if (num < minn) {
				minn = num;
				x1 = ans[i];
				y1 = ans[i+1];
			}
			if (num > maxx) {
				maxx = num;
				x2 = ans[i];
				y2 = ans[i+1];
			}
		}
		if (!maxx) puts("There are no adjacent primes.");
		else printf("%d,%d are closest, %d,%d are most distant.\n", x1, y1, x2, y2);
	}
	return 0;
}

阶乘分解

#include<bits/stdc++.h>
using namespace std;
inline long long read(){
    long long num=0;int z=1;char c=getchar();
    if(c=='-') z=-1;
    while((c<'0'||c>'9')&&c!='-') c=getchar();
    if(c=='-') z=-1,c=getchar();
    while(c>='0'&&c<='9') num=(num<<1)+(num<<3)+(c^48),c=getchar();
    return z*num;
}
#define _for(i,a,b) for(int i=a;i<=b;i++)
int n;
const int maxx=1000005;
int v[maxx],s[maxx];
int tot;
void find()
{
    _for(i,2,n)
    {
        if(v[i])continue;
        ++tot;
        s[tot]=i;
        _for(j,2,n/i)v[i*j]=1;
    }
} 
void work()
{
    _for(i,1,tot)
    {
        int p=s[i];int c=0;
        for(int j=n;j;j=j/p)
        {
            c+=j/p;
        }
        printf("%d %d\n",s[i],c);
    }
}
int main(){
    n=read();
    find();
    work();
    return 0;
}