UVA10140 Prime Distance

嘟嘟嘟

 

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。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxp = 5e5 + 5;
const int maxn = 1e6 + 5;
inline ll read()
{
    ll ans = 0;
    char ch = getchar(), last = ' ';
    while(!isdigit(ch)) {last = ch; ch = getchar();}
    while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();}
    if(last == '-') ans = -ans;
    return ans;
}
inline void write(ll x)
{
    if(x < 0) x = -x, putchar('-');
    if(x >= 10) write(x / 10);
    putchar(x % 10 + '0');
}

ll L, R;
int prime[maxp], v[maxp];
bool vis[maxn];

void init()
{
    for(int i = 2; i < maxp; ++i)
    {
        if(!v[i]) v[i] = i, prime[++prime[0]] = i;
        for(int j = 1; j <= prime[0] && prime[j] * i <= maxp; ++j)
        {
            if(prime[j] > v[i]) break;
            v[i * prime[j]] = prime[j];        
        }
    }
}

int main()
{
    init();
    while(scanf("%lld%lld", &L, &R) != EOF)
    {
        Mem(vis, 0); if(L == 1) vis[0] = 1;
        for(int i = 1; (ll)prime[i] * (ll)prime[i] <= R && i <= prime[0]; ++i)
        {
            for(int j = (L + prime[i] - 1) / prime[i]; j <= R / prime[i]; ++j)
                if(j > 1) vis[j * prime[i] - L] = 1;
        }
        ll Min1 = 0, Min2 = 0, Max1 = 0, Max2 = 0, now1 = 0, now2 = 0;
        for(int i = 0; i <= R - L; ++i)
        {
            if(!vis[i]) 
            {
                now1 = now2; now2 = i + L;
                if(now1)
                {
                    if(!Min1 || now2 - now1 < Min2 - Min1) Min1 = now1, Min2 = now2;
                    if(!Max1 || now2 - now1 > Max2 - Max1) Max1 = now1, Max2 = now2;
                }
            }
        }
        if(!Min1) printf("There are no adjacent primes.
");
        else printf("%lld,%lld are closest, %lld,%lld are most distant.
", Min1, Min2, Max1, Max2);
    }
    return 0;
}