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

Submit Status Practice UVA 10200

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;
}