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  • uva 11105

    版权声明:本文为博主原创文章,未经博主同意不得转载。

    https://blog.csdn.net/u011328934/article/details/36644069

    题目链接:uva 11105 - Semi-prime H-numbers

    题目大意:H-number为4k+1(k为非负数),H-composites为因子中含有H-number(不包含自己本身)的数。反之久是H-prime,给定n。求有多少H-composites。

    解题思路:首先用筛选法求出范围内的H-prime。然后枚举两个推断乘积是否在范围内

    #include <cstdio>
    #include <cstring>
    
    const int maxn = 1e6+5;
    typedef long long ll;
    
    int v[maxn], prime[maxn], cp;
    
    void primeHtable(int n) {
        cp = 0;
        memset(v, 0, sizeof(v));
    
        for (int i = 5; i < n; i += 4) {
            if (v[i])
                continue;
    
            prime[cp++] = i;
            for (int j = i * 2; j < n; j += i)
                v[j] = 1;
        }
    }
    
    int solve (int n) {
        int ans = 0;
        memset(v, 0, sizeof(v));
    
        for (int i = 0; prime[i] < n && i < cp; i++) {
            if ((ll)prime[i] * prime[i] > n)
                break;
    
            for (int j = i; prime[j] < n && j < cp; j++) {
    
                ll u = (ll)prime[i] * prime[j];
                if (u > n)
                    break;
    
    
                if (v[u])
                    continue;
                ans++;
                v[u] = 1;
            }
        }
        return ans;
    }
    
    int main () {
        primeHtable(maxn);
        int n;
        while (scanf("%d", &n) == 1 && n) {
            printf("%d %d
    ", n, solve(n));
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/ldxsuanfa/p/10720144.html
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