POJ 2262 Goldbach's Conjecture
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Goldbach's Conjecture
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 38917 Accepted: 14906
Description
In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture:
For example:
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.
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.
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
820420
Sample Output
8 = 3 + 520 = 3 + 1742 = 5 + 37
Source
Ulm Local 1998
解题思路:
和刚才上面有一道题很像,但是这道题的数据变大了 1000000,第一次用了O(n^2) 果断TLE了,其实不用这么SB的算法,只要在这个筛选出来的素数表中把
n-i和i的项找出来,分别判断两者是否同时为1就可以了,如果两者同时为1,那么,就可以认为找到了,这组我们需要的解,如果两者不同时为1,那么就continue.如果i的变化范围超过了n/2都没有找到合适的解,那么就可以认为无解了.
代码:
# include<cstdio># include<iostream>using namespace std;# define MAX 1000000int a[MAX];int b[MAX];void dabiao() { for ( int i = 2;i <= MAX;i++ ) a[i] = 1; for ( int i = 2;i <= MAX;i++ ) { if ( a[i] == 1 ) { for ( int j = 2*i;j <= MAX;j+=i ) { if ( j%i == 0 ) a[j] = 0; } } } }int main(void){ dabiao(); int n; while ( cin>>n ) { int x = 0; int y = 0; if ( n == 0 ) break; int i; for ( i = 3;i <= n/2;i++ ) { if ( a[i] == 1 && a[n-i] == 1 ) { printf("%d = %d + %d\n", n, i, n-i); break; } } if ( i > n/2 ) { printf("Goldbach's conjecture is wrong.\n"); } } // } return 0;}
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