- 题意：给L，R（1<=L<=R<=2^31，R-L<=10^6），找到[范围L,R]之间相差最大的相邻素数。
- 题解：
- 首先注意L,R可能爆int
- 然后直接筛出2~sqrt(R)之间的所有素数
- 再用这些素数筛出[L,R]中的所有素数
- 找相邻差最大即可
- 代码：
//#include<bits/stdc++.h>
#include<algorithm>
#include <iostream>
#include <fstream>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <vector>
#include <string>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define P(a,b,c) make_pair(a,make_pair(b,c))
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define CLR(vis) memset(vis,0,sizeof(vis))
#define MST(vis,pos) memset(vis,pos,sizeof(vis))
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef pair<int,pair<int,int> >pii;
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1000000007;
const int INF = 0x3f3f3f3f;
ll gcd(ll a, ll b) {
return b ? gcd(b, a%b) : a;
}
template<class T>inline void gmax(T &A, T B){
(A<B)&&(A=B);//if(B>A)A=B;
}
template<class T>inline void gmin(T &A, T B){
(A>B)&&(A=B);//if(B<A)A=B;
}
template <class T>
inline bool scan_d(T &ret) {
char c;
int sgn;
if(c=getchar(),c==EOF) return 0; //EOF
while(c!='?'&&(c<'0'||c>'9')) c=getchar();
sgn=(c=='?')?-1:1;
ret=(c=='?')?0:(c-'0');
while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
ret*=sgn;
return 1;
}
inline void outt(int x) {
if(x>9) outt(x/10);
putchar(x%10+'0');
}
const int MAXN = 1e6 + 10;
const int MAX = 1e5;
int prime[MAX], tag[MAX], vis[MAXN], tot;
void get_prime(void){
for(int i = 2; i < MAX; i++){
if(!tag[i]){
prime[tot++] = i;
for(int j = 2; j * i < MAX; j++){
tag[j * i] = 1;
}
}
}
}
ll Max(ll a, ll b){
return a > b ? a : b;
}
int main(void){
get_prime();
ll l, r;
while(~scanf("%lld%lld", &l, &r)){
memset(vis, 0, sizeof(vis));
for(int i = 0; i < tot; i++){
ll a = (l + prime[i] - 1) / prime[i];
ll b = r / prime[i];
for(int j = Max(2, a); j <= b; j++){ // 筛[l, r]内的合数
vis[prime[i] * j - l] = 1; //减个l方便标记,输出答案时加回去即可
}
}
if(l == 1) vis[0] = 1; // 注意这个1并不是素数
ll cnt = -1, sol1 = MAXN, sol2 = 0, x1, y1, x2, y2;
for(int i = 0; i <= r - l; i++){
if(vis[i] == 0){
if(cnt != -1){
if(sol1 > i - cnt){
x1 = cnt;
y1 = i;
sol1 = i - cnt;
}
if(sol2 < i - cnt){
x2 = cnt;
y2 = i;
sol2 = i - cnt;
}
}
cnt = i;
}
}
if(sol2 == 0) puts("There are no adjacent primes.");
else printf("%lld,%lld are closest, %lld,%lld are most distant.\n", x1 + l, y1 + l, x2 + l, y2 + l);
}
return 0;
}
撰写