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  • POJ2262 Goldbach's Conjecture

    2262:Goldbach's Conjecture


    题目分析:

    一百万个数,开辟一个数组用于标识是否是素数,用bool型表示,总内存小于1M,然后查表即可。建立素数表,对每个素数,把该素数的所有倍数所在的位置都置为false。

    代码如下:

    1048kB 50ms688 B

    #include <stdio.h>
    int main(){
        bool primeMap[1000000];
        for(int i=1; i<1000000; i++){
            primeMap[i] = true;
        }
        for(int i=2; i<500000; i++){
            if(primeMap[i] == true){
                for(int j = 2*i; j < 1000000; j+=i){
                    primeMap[j] = false;
                }
            }
        }
        int n;
        while(scanf("%d",&n) && n!=0){
            int ia = 2;
            for(; ia<=n/2; ia++){
                if(primeMap[ia] == true && primeMap[n - ia] == true){
                    printf("%d = %d + %d\n",n, ia, n-ia);
                    break;
                }
            }
            if(ia == n/2 + 1){
                printf("Goldbach's conjecture is wrong.\n");
            }
        }
        return 0;
    }
    题目:

    时间限制: 1000ms 内存限制: 65536kB
    描述
    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.
    输入
    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.
    输出
    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."
    样例输入
    8
    20
    42
    0
    样例输出
    8 = 3 + 5
    20 = 3 + 17
    42 = 5 + 37

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  • 原文地址:https://www.cnblogs.com/zhuyuanhao/p/3262879.html
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