UVA10140 Prime Distance
L, R很大,所以哪怕用线性筛也会T飞的,但是看到R - L不太大,因此我们可以先筛出[1, √R]所有素数,然后用这些数筛出[L, R]的所有合数。具体做法是对于每一个pi,筛去所有的 i * pi (⌈L / i⌉ <= i <= ⌊R / i⌋)。然后遍历[L, R]的所有素数更新答案即可。
要注意的是1不是质数,然后如果素数表拿int存的话,prime[i] * prime[i]要强制转换成long long ,否则爆了后会无限循环然后RE。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cctype> 8 #include<vector> 9 #include<stack> 10 #include<queue> 11 using namespace std; 12 #define enter puts("") 13 #define space putchar(' ') 14 #define Mem(a, x) memset(a, x, sizeof(a)) 15 #define rg register 16 typedef long long ll; 17 typedef double db; 18 const int INF = 0x3f3f3f3f; 19 const db eps = 1e-8; 20 const int maxp = 5e5 + 5; 21 const int maxn = 1e6 + 5; 22 inline ll read() 23 { 24 ll ans = 0; 25 char ch = getchar(), last = ' '; 26 while(!isdigit(ch)) {last = ch; ch = getchar();} 27 while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();} 28 if(last == '-') ans = -ans; 29 return ans; 30 } 31 inline void write(ll x) 32 { 33 if(x < 0) x = -x, putchar('-'); 34 if(x >= 10) write(x / 10); 35 putchar(x % 10 + '0'); 36 } 37 38 ll L, R; 39 int prime[maxp], v[maxp]; 40 bool vis[maxn]; 41 42 void init() 43 { 44 for(int i = 2; i < maxp; ++i) 45 { 46 if(!v[i]) v[i] = i, prime[++prime[0]] = i; 47 for(int j = 1; j <= prime[0] && prime[j] * i <= maxp; ++j) 48 { 49 if(prime[j] > v[i]) break; 50 v[i * prime[j]] = prime[j]; 51 } 52 } 53 } 54 55 int main() 56 { 57 init(); 58 while(scanf("%lld%lld", &L, &R) != EOF) 59 { 60 Mem(vis, 0); if(L == 1) vis[0] = 1; 61 for(int i = 1; (ll)prime[i] * (ll)prime[i] <= R && i <= prime[0]; ++i) 62 { 63 for(int j = (L + prime[i] - 1) / prime[i]; j <= R / prime[i]; ++j) 64 if(j > 1) vis[j * prime[i] - L] = 1; 65 } 66 ll Min1 = 0, Min2 = 0, Max1 = 0, Max2 = 0, now1 = 0, now2 = 0; 67 for(int i = 0; i <= R - L; ++i) 68 { 69 if(!vis[i]) 70 { 71 now1 = now2; now2 = i + L; 72 if(now1) 73 { 74 if(!Min1 || now2 - now1 < Min2 - Min1) Min1 = now1, Min2 = now2; 75 if(!Max1 || now2 - now1 > Max2 - Max1) Max1 = now1, Max2 = now2; 76 } 77 } 78 } 79 if(!Min1) printf("There are no adjacent primes.\n"); 80 else printf("%lld,%lld are closest, %lld,%lld are most distant.\n", Min1, Min2, Max1, Max2); 81 } 82 return 0; 83 }