POJ 2262 Goldbach's Conjecture
2013-07-25 13:01
323 查看
[align=center]Goldbach's Conjecture[/align]
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
Sample Output
Source
Ulm Local 1998
题意:
将一个数拆成两个奇质数的和
代码:
思路:
额 水。。
在刷表的时候i和j为long long
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 33638 | Accepted: 12903 |
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
Ulm Local 1998
题意:
将一个数拆成两个奇质数的和
代码:
#include<cstdio> #include<cstdlib> #include<cstring> #define MAX 1000000 int p[MAX]; int main() { memset(p,0,sizeof(p)); p[0]=p[1]=1; long long i,j; for(i=2;i<=MAX;i++) if(!p[i]) { for(j=i*i;j<=MAX;j+=i) p[j]=1; if(i%2==0) p[i]=1; } //刷奇质数表 int n; while(scanf("%d",&n)!=EOF && n) { int k; int flag=0; for(k=3;k<n;k++) if(!p[k]) if(!p[n-k]) { flag=1; break; } if(flag==1) printf("%d = %d + %d\n",n,k,n-k); else printf("Goldbach's conjecture is wrong.\n"); } return 0; }
思路:
额 水。。
在刷表的时候i和j为long long
相关文章推荐
- poj2262 Goldbach's Conjecture
- POJ 2262 Goldbach's Conjecture(素数表分解质数)
- POJ-2262 Goldbach's Conjecture
- POJ 2262 Goldbach's Conjecture
- POJ 2262 Goldbach's Conjecture 哥德巴赫猜想
- poj 2262 Goldbach's Conjecture 素数筛
- POJ 2262 Goldbach's Conjecture [暴力]
- poj-2262-Goldbach's Conjecture
- POJ 2262 Goldbach's Conjecture(素数表)
- POJ 2262 Goldbach's Conjecture 简单的素数
- POJ 2262 Goldbach's Conjecture
- poj 2262 Goldbach's Conjecture 【素数筛法】
- poj_2262 Goldbach's Conjecture
- poj 2262 Goldbach's Conjecture 【素数筛】
- POJ 2262 / UVa 543 - Goldbach's Conjecture (哥德巴赫猜想)
- POJ 2262 Goldbach's Conjecture(素数)
- POJ 2262 Goldbach's Conjecture
- POJ 2262 Goldbach's Conjecture (水题)
- POJ 刷题系列:2262. Goldbach's Conjecture
- poj-2262 Goldbach's Conjecture