POJ 2689 Prime Distance 素数筛法
Prime Distance
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 17075 Accepted: 4556
Description
The branch of mathematics called number theory is about properties of numbers. One of the areas that has captured the interest of number theoreticians for thousands of years is the question of primality. A prime number is a number that is has no proper factors (it is only evenly divisible by 1 and itself). The first prime numbers are 2,3,5,7 but they quickly become less frequent. One of the interesting questions is how dense they are in various ranges. Adjacent primes are two numbers that are both primes, but there are no other prime numbers between the adjacent primes. For example, 2,3 are the only adjacent primes that are also adjacent numbers.
Your program is given 2 numbers: L and U (1<=L< U<=2,147,483,647), and you are to find the two adjacent primes C1 and C2 (L<=C1< C2<=U) that are closest (i.e. C2-C1 is the minimum). If there are other pairs that are the same distance apart, use the first pair. You are also to find the two adjacent primes D1 and D2 (L<=D1< D2<=U) where D1 and D2 are as distant from each other as possible (again choosing the first pair if there is a tie).
Input
Each line of input will contain two positive integers, L and U, with L < U. The difference between L and U will not exceed 1,000,000.
Output
For each L and U, the output will either be the statement that there are no adjacent primes (because there are less than two primes between the two given numbers) or a line giving the two pairs of adjacent primes.
Sample Input
2 17
14 17
Sample Output
2,3 are closest, 7,11 are most distant.
There are no adjacent primes.
Source
Waterloo local 1998.10.17题目给出的数据范围很大 但是U-L不超过1e6
可以通过素数筛法把1e5以内的素数先求出
再用1e5内的素数对[L,U]筛(去掉[L,U]中1e5内的素数的倍数)
#include
#includeif(isprime[i]){
prime.push_back(i);
for(ll j=(ll)i*i;jfalse;
}
}
vector<int>prime2;//[L,U]范围内的素数
const int maxlen=1000000+5;
bool isprime2[maxlen];
void doPrime(ll L,ll U) {
prime2.clear();
fill(isprime2, isprime2 + maxlen, true);
for (int i = L,end=min(U,N-1); i <= end; ++i)//[L , min(N-1,U)]
isprime2[i - L] = isprime[i];
ll tmpL=L;
L = max(L, N);
if (L <= U) {
for (int i = 0; i < prime.size(); ++i) {
ll pr = prime[i];
if (pr * pr > U)
break;
ll s = L / pr,
e = U / pr;
while (s <= e) {//筛去[L,U]范围内pr的倍数
ll t = s * pr;
if (s > 1 && t >= L && t <= U)
isprime2[s * pr - L] = false;
++s;
}
}
}
L=tmpL;
for(ll i=L;i<=U;++i)
if(isprime2[i-L])
prime2.push_back(i);
}
void slove(ll L,ll U){
doPrime(L,U);//求出[L,U]内的素数
if(prime2.size()<2){
puts("There are no adjacent primes.");
return ;
}
int minDt=INF;
int minA,minB;
int maxDt=0;
int maxA,maxB;
for(int i=0;i1;++i){//找最短距离和最远距离
int j=i+1;
if(prime2[j]-prime2[i]if(prime2[j]-prime2[i]>maxDt){
maxDt=prime2[j]-prime2[i];
maxA=prime2[i],maxB=prime2[j];
}
}
printf("%d,%d are closest, %d,%d are most distant.\n",
minA,minB,maxA,maxB);
}
int main()
{
//freopen("/home/lu/文档/r.txt","r",stdin);
//freopen("/home/lu/文档/w.txt","w",stdout);
iniPrime();//提前筛出<1e5的素数
ll L,U;
while(~scanf("%lld%lld",&L,&U)){
slove(L,U);
}
return 0;
}