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  • poj 2262 筛法求素数(巧妙利用数组下标!)

    Goldbach's Conjecture
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 41582   Accepted: 15923

    Description

    In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture:
    Every even number greater than 4 can be
    written as the sum of two odd prime numbers.

    For example:
    8 = 3 + 5. Both 3 and 5 are odd prime numbers.
    20 = 3 + 17 = 7 + 13.
    42 = 5 + 37 = 11 + 31 = 13 + 29 = 19 + 23.

    Today it is still unproven whether the conjecture is right. (Oh wait, I have the proof of course, but it is too long to write it on the margin of this page.)
    Anyway, your task is now to verify Goldbach's conjecture for all even numbers less than a million.

    Input

    The input will contain one or more test cases.
    Each test case consists of one even integer n with 6 <= n < 1000000.
    Input will be terminated by a value of 0 for n.

    Output

    For each test case, print one line of the form n = a + b, where a and b are odd primes. Numbers and operators should be separated by exactly one blank like in the sample output below. If there is more than one pair of odd primes adding up to n, choose the pair where the difference b - a is maximized. If there is no such pair, print a line saying "Goldbach's conjecture is wrong."

    Sample Input

    8
    20
    42
    0
    

    Sample Output

    8 = 3 + 5
    20 = 3 + 17
    42 = 5 + 37
    

    Source

    心得:自己写的筛法时间复杂度是O(n^2),利用数组下标省去一个循环后变成了O(n);妙不可言!!!
    #include <iostream>
    #include<cstdio>
    #include<cstring>
    using namespace std;
    #define N 1000000
    int n,a[N];
    bool pr[N];
    void pdd()
    {
        int i,j;
        for(i=0;i<1000000;i++) a[i]=1;
        a[0]=0; a[1]=0;
        for(i=2;i<1000000;i++)
        {
            if(a[i]==1)
            {
                for(j=i*2;j<1000000;j+=i) a[j]=0;
            }
        }
    }
    int main()
    {
        int num;
        int i,flag;
        pdd();
        while(scanf("%d",&num)!=EOF&&num!=0)
        {
            for(flag=0,i=2;i<num;i++)
            {
                if(a[i]==1&&a[num-i]==1)
                {
                    printf("%d = %d + %d
    ",num,i,num-i);
                    flag=1;
                    break;
                }
            }
            if(flag==0)
                printf("Goldbach's conjecture is wrong.
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/yoyo-sincerely/p/5024676.html
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