Prime counting,sum of primes,leetcode,dp.

主要参考 http://www.zhihu.com/question/29580448/answer/45218281

https://projecteuler.net/thread=10;page=5   Lucy_Hedgehog 的代码

如果不用map，自己写hash函数会更快

// 计算 Π(n-1)class Solution {
public:
    int countPrimes(int n) {
        n--;
        if(n<=1) return 0;
        int r = int(sqrt(n));
        map<int ,int> hmp;
        for(int i = 1;i<=r;i++) hmp[n/i] = n/i - 1;
        for(int i = n/r -1;i>0;i--) hmp[i] = i - 1;
        for(int p = 2;p<=r;p++){
            if(hmp[p]>hmp[p-1]){
                int sp = hmp[p-1];
                int p2 = p*p;
                for(int i = 1;i<=r;i++){
                    int t = n/i;
                    if(t<p2) break;
                    hmp[t] -= hmp[t/p] - sp;
                }
                for(int i = n/r -1;i>0;i--){
                    if(i<p2) break;
                    hmp[i] -= hmp[i/p] - sp;
                }
            }
        }
    return hmp[n];
    }
}；

//计算 2....n之间的素数和long long  P10(long long n){
    long long r = long long(sqrt(n));
    hash_map<long long,long long> hmp;
    for(long long i = 1;i<=r;i++){
        long long t = n/i;
        hmp[t] = t*(t+1)/2 - 1;
    }
    for(long long i = n/r -1;i>0;i--){
        long long t = i;
        hmp[t] = t*(t+1)/2 - 1;
    }
    for(long long p = 2;p<=r;p++){
        if(hmp[p]>hmp[p-1]){
            long long sp = hmp[p-1];
            long long p2 = p*p;
            for(long long i = 1;i<=r;i++){
                long long t = n/i;
                if(t<p2) break;
                hmp[t] -= p*(hmp[t/p] - sp);
            }
            for(long long i = n/r -1;i>0;i--){
                long long t = i;
                if(t<p2) break;
                hmp[t] -= p*(hmp[t/p] - sp);
            }
        }
    }
    return hmp[n];
}