整理的算法模板合集: ACM模板
试除法判定质数
bool is_prime(int x)
{
if (x < 2) return false;
for (int i = 2; i <= x / i; i ++ )
if (x % i == 0)
return false;
return true;
}
线性筛法
int primes[N], cnt;
bool vis[N];
void init(int x){
vis[1] = vis[0] = 1;
for(int i = 2; i <= x; ++ i){
if(!vis[i])primes[cnt ++ ] = i;
for(int j = 0; primes[j] * i <= x; ++ j){
vis[primes[j] * i] = true;
if(i % primes[j] == 0)break;
}
}
}
二次筛法
/*AcWing 196. 质数距离
整数数列1~n,给出区间[L,R] (L,R<=2^31,R-L<=1e6) 求区间内相邻质数距离的最大值和最小值*/
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 1000010;
int primes[N], cnt;
bool st[N];
void init(int n)
{
memset(st, 0, sizeof st);
cnt = 0;
for (int i = 2; i <= n; i ++ )
{
if (!st[i]) primes[cnt ++ ] = i;
for (int j = 0; primes[j] * i <= n; j ++ )
{
st[i * primes[j]] = true;
if (i % primes[j] == 0) break;
}
}
}
int main()
{
int l, r;
while (cin >> l >> r)
{
init(50000);
memset(st, 0, sizeof st);
for (int i = 0; i < cnt; i ++ )
{
LL p = primes[i];
for (LL j = max(p * 2, (l + p - 1) / p * p); j <= r; j += p)
st[j - l] = true;
}
cnt = 0;
for (int i = 0; i <= r - l; i ++ )
if (!st[i] && i + l >= 2)
primes[cnt ++ ] = i + l;
if (cnt < 2) puts("There are no adjacent primes.");
else
{
int minp = 0, maxp = 0;
for (int i = 0; i + 1 < cnt; i ++ )
{
int d = primes[i + 1] - primes[i];
if (d < primes[minp + 1] - primes[minp]) minp = i;
if (d > primes[maxp + 1] - primes[maxp]) maxp = i;
}
printf("%d,%d are closest, %d,%d are most distant.\n",
primes[minp], primes[minp + 1],
primes[maxp], primes[maxp + 1]);
}
}
return 0;
}