这题……一开始没想到 后来
题意就是求区间素数对最大和最小距离
发现必须处理所有素数
复杂度要求是O(n)～O(nlgn)
考虑分开求质数和合数 其实就是筛法筛合数
最后遍历一遍找最大最小值即可
然后这个方法筛素数到R½就可以了
也就是50000（WA是写的40000……）
Time cost:55min
Code:
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<cmath> 5 #include<queue> 6 #include<vector> 7 #define ms(a,b) memset(a,b,sizeof a) 8 #define rep(i,a,n) for(int i = a;i <= n;i++) 9 #define per(i,n,a) for(int i = n;i >= a;i--) 10 #define inf 2147483647 11 using namespace std; 12 typedef long long ll; 13 ll read() { 14 ll as = 0,fu = 1; 15 char c = getchar(); 16 while(c < '0' || c > '9') { 17 if(c == '-') fu = -1; 18 c = getchar(); 19 } 20 while(c >= '0' && c <= '9') { 21 as = as * 10 + c - '0'; 22 c = getchar(); 23 } 24 return as * fu; 25 } 26 const int N = 100005; 27 //head 28 ll l,r; 29 ll mindiv[N],prime[N]; 30 int top; 31 void initprime(int n) { 32 rep(i,2,n) { 33 if(!mindiv[i]) prime[++top] = i; 34 for(int j = 1;j <= top && i * prime[j] <= n;j++) { 35 mindiv[prime[j] * i] = prime[j]; 36 if(i % prime[j] == 0) break; 37 } 38 } 39 } 40 41 int flag[N<<4],tot; 42 43 int main() { 44 initprime(50000); 45 while(~scanf("%lld%lld",&l,&r)) { 46 tot++; 47 if(l == 1) l++; 48 rep(i,1,top) { 49 if(prime[i] * prime[i] > r) break; 50 rep(j,(l+prime[i]-1)/prime[i],r/prime[i]) 51 if(j > 1) flag[j * prime[i] - l] = tot; 52 } 53 ll minn = 10000007,maxx = 0; 54 ll minid[2],maxid[2]; 55 ll cur = 0,curr = 0; 56 rep(i,l,r) { 57 if(flag[i-l] == tot) continue; 58 cur = curr,curr = i; 59 if(!cur) continue; 60 if(curr - cur < minn) minn = curr - cur,minid[0] = cur,minid[1] = curr; 61 if(curr - cur > maxx) maxx = curr - cur,maxid[0] = cur,maxid[1] = curr; 62 } 63 if(!maxx) puts("There are no adjacent primes."); 64 else printf("%lld,%lld are closest, %lld,%lld are most distant. ",minid[0],minid[1],maxid[0],maxid[1]); 65 } 66 return 0; 67 }