Sample Input
2 17 14 17
Sample Output
2,3 are closest, 7,11 are most distant. There are no adjacent primes.
找出给定范围内,距离最远和最近的素数。(不停超时 - -)
给的上界很大,所以全处理肯定不行。 先处理sqrt(2147483647)。
然后再在l 与 r之间筛选素数。
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <cmath> #define N 10100 typedef long long ll; using namespace std; int n,m; int f[1000500]; int q[1000500]; int t[100050]; int tot; void prim() { ll i,j; for(i = 2; i <= 100050; i++) q[i] = 1; for(i= 2,tot = 0; i <= 100050; i++) { if(q[i]) { t[tot++] = i; for(j = i*2; j <= 100050; j+=i) q[j] = 0; } } } int main() { int l,r; while(scanf("%d%d",&l,&r)!= EOF) { prim(); if(l == 1) l = 2; memset(f,0,sizeof(f)); for(int i =0; i < tot; i++) { int a = (l-1)/t[i]+1; int b = r /t[i]; for(int j = a; j <= b; j++) //类似上面prim()的方法 if(j > 1) f[j*t[i]-l] = 1; } int Max = -1; int Min = 1000000000; int tmp = -1; int maxl,maxr,minl,minr; for(int i = 0; i <= r-l; i++) { if(!f[i]) { if(tmp>=0 && i - tmp> Max) { Max = i-tmp; maxl = tmp+l; maxr = i+l; } if(tmp>=0 && i - tmp< Min) { Min = i-tmp; minl = tmp+l; minr = i+l; } tmp = i; } } if(Max == -1) printf("There are no adjacent primes. "); else { printf("%d,%d are closest, %d,%d are most distant. ",minl,minr,maxl,maxr); } } return 0; }