UVA 10200 记忆打表，素数筛，浮点误差

UVA 10200 区间预处理，浮点误差

W - Prime Time

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

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Description

 

题意：测试找素数函数f(n)=n^2+n+41在区间n<-[a,b]时，找到素数的成功率。

思路：区间的素数个数，打表预处理即可。输出答案时四舍五入，需注意浮点误差，四舍五入时加上EPS。

#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>

#define ll long long

#define REP(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)

#define REPP(i,a,b,t) for(int (i)=(a);(i)<=(b);(i)+=(t))

#define PII pair<int,int>

#define MP make_pair

#define PB push_back

#define RI(x) scanf("%d",&(x))

#define RLL(x) scanf("%lld",&(x))

#define RI64(x) scanf("%I64d",&(x))

#define DRI(x) int x;scanf("%d",&(x))

#define DRLL(x) ll x;scanf("%lld",&(x))

#define DRI64(x) llx;scanf("%I64d",&(x))

#define MS0(a) memset((a),0,sizeof((a)))

#define MS1(a) memset((a),0,sizeof((a)))

#define MS(a,b) memset((a),(b),sizeof((a)))



using namespace std;



const int maxn=10000100;

const int INF=(1<<29);

const double EPS=0.0000000001;

const double Pi=acos(-1.0);



ll a,b;

bool isprime[maxn*10];

vector<int> prime;

ll cnt[maxn];



void getPrime()

{

    MS(isprime,1);

    isprime[1]=0;

    REP(i,2,maxn-1){

        if(!isprime[i]) continue;

        REPP(j,i*2,maxn-1,i) isprime[j]=0;

    }

    REP(i,1,maxn-1) if(isprime[i]) prime.PB(i);

}



ll f(ll n)

{

    return n*n+n+41;

}



bool isP(ll n)

{

    REP(i,0,prime.size()){

        if(prime[i]*1LL*prime[i]>n||prime[i]*1LL*prime[i]<=0) break;

        if(n%prime[i]==0) return 0;

    }

    return 1;

}



void getCnt()

{

    MS0(cnt);

    cnt[0]=1;

    REP(i,1,10010){

        ll x=f(i);

        if(x<maxn){

            if(isprime[x]) cnt[i]=cnt[i-1]+1;

            else cnt[i]=cnt[i-1];

        }

        else{

            if(isP(x)) cnt[i]=cnt[i-1]+1;

            else cnt[i]=cnt[i-1];

        }

    }

}



int main()

{

    getPrime();

    getCnt();

    //cout<<f(10000)<<endl;

    while(cin>>a>>b){

        printf("%.2f\n",(cnt[b]-(a?cnt[a-1]:0))*100.0/(b-a+1)+EPS);

    }

    return 0;

}

View Code