light_oj 1197 区间素数筛

light_oj 1197 区间素数筛

M - Help Hanzo

Time Limit:2000MS      Memory Limit:32768KB      64bit IO Format:%lld & %llu

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Description

Amakusa, the evil spiritual leader has captured the beautiful princess Nakururu. The reason behind this is he had a little problem with Hanzo Hattori, the best ninja and the love of Nakururu. After hearing the news Hanzo got extremely angry. But he is clever and smart, so, he kept himself cool and made a plan to face Amakusa.

Before reaching Amakusa's castle, Hanzo has to pass some territories. The territories are numbered as a, a+1, a+2, a+3 ... b. But not all the territories are safe for Hanzo because there can be other fighters waiting for him. Actually he is not afraid of them, but as he is facing Amakusa, he has to save his stamina as much as possible.

He calculated that the territories which are primes are safe for him. Now given a and b he needs to know how many territories are safe for him. But he is busy with other plans, so he hired you to solve this small problem!

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case contains a line containing two integers a and b (1 ≤ a ≤ b < 2³¹, b - a ≤ 100000).

Output

For each case, print the case number and the number of safe territories.

Sample Input

3

2 36

3 73

3 11

Sample Output

Case 1: 11

Case 2: 20

Case 3: 4

题意：给定a,b，求区间[a,b]中的素数个数。

思路：先常规的素数筛法筛出[1,10^6]的素数，再根据[1,10^6]的素数筛出[a,b]中的素数。

一开始用map超时了，改成set也超时了。。。最后改成常规的bool数组就过了，看来map和set的常数还是有点大。。像这种下标较大但下标范围较小的还是用常规的数组使下标加上一个常数比较好。

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<algorithm>

#include<vector>

#include<stack>

#include<queue>

#include<set>

#include<map>

#include<string>

#include<math.h>

#include<cctype>



using namespace std;



typedef long long ll;

const int maxn=1000100;

const int INF=(1<<29);

const double EPS=0.0000000001;

const double Pi=acos(-1.0);



int T;

ll a,b;

bool isprime[maxn];

vector<ll> prime;

bool isP[maxn];



void play_prime()

{

    memset(isprime,1,sizeof(isprime));

    isprime[1]=0;

    for(int i=2;i<maxn;i++){

        if(!isprime[i]) continue;

        for(int j=i*2;j<maxn;j+=i){

            isprime[j]=0;

        }

    }

    for(int i=1;i<maxn;i++){

        if(isprime[i]) prime.push_back(i);

    }

}



int main()

{

    cin>>T;

    int tag=1;

    play_prime();

    while(T--){

        scanf("%lld%lld",&a,&b);

        if(a<=1) a=2;

        memset(isP,1,sizeof(isP));

        for(int i=0;i<prime.size()&&prime[i]*prime[i]<=b;i++){

            ll p=prime[i];

            for(ll j=a+(p-a%p)%p;j<=b;j+=p){

                if(j!=p) isP[j-a]=0;

            }

        }

        int ans=0;

        for(ll i=a;i<=b;i++){

            if(isP[i-a]) ans++;

        }

        printf("Case %d: %d\n",tag++,ans);

    }

    return 0;

}

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