• Primes on Interval(二分 + 素数打表)


    Primes on Interval
    Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u
    Appoint description: 

    Description

    You've decided to carry out a survey in the theory of prime numbers. Let us remind you that a prime number is a positive integer that has exactly two distinct positive integer divisors.

    Consider positive integers a, a + 1, ..., b(a ≤ b). You want to find the minimum integer l(1 ≤ l ≤ b - a + 1) such that for any integer x(a ≤ x ≤ b - l + 1) among l integers x, x + 1, ..., x + l - 1 there are at least k prime numbers.

    Find and print the required minimum l. If no value l meets the described limitations, print -1.

    Input

    A single line contains three space-separated integers a, b, k (1 ≤ a, b, k ≤ 106a ≤ b).

    Output

    In a single line print a single integer — the required minimum l. If there's no solution, print -1.

    Sample Input

    Input
    2 4 2
    Output
    3
    Input
    6 13 1
    Output
    4
    Input
    1 4 3
    Output
    -1
    题意:
    求最小的l使 x, x + 1, ..., x + l - 1 there are at least k prime numbers;
    简单二分;
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    using namespace std;
    const int MAXN = 1e6 + 100;
    int dp[MAXN];
    int vis[MAXN];
    void db(){
        memset(vis, 0, sizeof(vis));
        vis[1] = 1;
        for(int i = 2; i <= sqrt(MAXN); i++){
            if(!vis[i]){
                for(int j = i * i; j < MAXN; j += i){
                    vis[j] = 1;
                }
            }
        }
        dp[0] = 0;
        for(int i = 1; i < MAXN; i++){
            dp[i] = dp[i - 1];
            if(!vis[i])dp[i]++;
        }
    }
    bool js(int l, int a, int b, int k){
        for(int i = a; i <= b - l + 1; i++){
            if(dp[i + l - 1] - dp[i - 1] < k)return false;
        }
        return true;
    }
    int erfen(int l, int r, int a, int b, int k){
        int mid, ans = -1;
        while(l <= r){
            mid = (l + r) >> 1;
            if(js(mid, a, b, k)){
                ans = mid;
                r = mid - 1;
            }
            else l = mid + 1;
        }
        return ans;
    }
    int main(){
        db();
        int a, b, k;
        while(~scanf("%d%d%d", &a, &b, &k)){
            printf("%d
    ", erfen(0, b - a + 1, a, b, k));
        }
        return 0;
    }
  • 相关阅读:
    #Leetcode# 219. Contains Duplicate II
    #Leetcode# 203. Remove Linked List Elements
    #Leetcode# 141. Linked List Cycle
    #Leetcode# 128. Longest Consecutive Sequence
    linux——shell解释
    Linux——互联网搜索引擎nbtscan是一个扫描WINDOWS网络NetBIOS信息的小工具
    Linux——网络端口的状态netstat、ifconfig
    连不上网的原因
    jquery基础笔记
    网址收藏
  • 原文地址:https://www.cnblogs.com/handsomecui/p/5440258.html
Copyright © 2020-2023  润新知