• POJ 2262


    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 + 520 = 3 + 1742 = 5 + 37

    题意:任意一个数字都可以写成两质数之和,输入一个数字,列出其与两质数之和的等式。

    考查素数筛的使用,珍爱生命,学会打表。

     1 #include <string>
     2 #include <iostream>
     3 #include <map>
     4 #include <cstring>
     5 #include <algorithm>
     6 #include <math.h>
     7 #include <stdio.h>
     8 #define N 1000001
     9 using namespace std;
    10 
    11 int main()
    12 {
    13     bool a[N];
    14     int n;
    15     a[0]=a[1]=1;
    16     for(int i=2;i<N;i++)
    17     {
    18             if(!a[i])
    19             for(int j=2*i;j<N;j+=i)
    20             {
    21                 a[j]=1;
    22             }
    23     }
    24     while(scanf("%d",&n)&&n)
    25     {
    26         for(int i=0;i<=n;i++)
    27         {
    28             if(!a[i]&&!a[n-i])
    29             {
    30                 printf("%d = %d + %d
    ",n,i,n-i);
    31                 break;
    32             }
    33         }
    34     }
    35     return 0;
    36 }
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  • 原文地址:https://www.cnblogs.com/greenaway07/p/10420018.html
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