HDU - 1397 Goldbach's Conjecture
来源:互联网 发布:手机淘宝能收藏店铺吗 编辑:程序博客网 时间:2024/05/26 07:29
Goldbach's Conjecture
Time Limit: 1000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u
Description
Goldbach's Conjecture: For any even number n greater than or equal to 4, there exists at least one pair of prime numbers p1 and p2 such that n = p1 + p2.
This conjecture has not been proved nor refused yet. No one is sure whether this conjecture actually holds. However, one can find such a pair of prime numbers, if any, for a given even number. The problem here is to write a program that reports the number of all the pairs of prime numbers satisfying the condition in the conjecture for a given even number.
A sequence of even numbers is given as input. Corresponding to each number, the program should output the number of pairs mentioned above. Notice that we are interested in the number of essentially different pairs and therefore you should not count (p1, p2) and (p2, p1) separately as two different pairs.
This conjecture has not been proved nor refused yet. No one is sure whether this conjecture actually holds. However, one can find such a pair of prime numbers, if any, for a given even number. The problem here is to write a program that reports the number of all the pairs of prime numbers satisfying the condition in the conjecture for a given even number.
A sequence of even numbers is given as input. Corresponding to each number, the program should output the number of pairs mentioned above. Notice that we are interested in the number of essentially different pairs and therefore you should not count (p1, p2) and (p2, p1) separately as two different pairs.
Input
An integer is given in each input line. You may assume that each integer is even, and is greater than or equal to 4 and less than 2^15. The end of the input is indicated by a number 0.
Output
Each output line should contain an integer number. No other characters should appear in the output.
Sample Input
610120
Sample Output
121
#include<iostream>#include<vector> //首先筛出素数,用vis标记每个数是不是素数,然后对于询问就for一遍看下每种拆分是不是两个都是素数。using namespace std;const int MAXN = 1000000; //注意要开的够大int vis[MAXN];void Prime() //素数筛{vis[0] = 1; vis[1] = 1;for (long long int i = 2; i*i <= MAXN; i++){if (!vis[i]){for (long long int j = i+i; j <= MAXN; j = j + i)vis[j] = 1;}}return;}int main(){Prime();long long int n;while (cin >> n&&n){long long int sum = 0;for (long long int i = 2; i <=n/2; i++){if (!vis[i] && !vis[n - i]) sum++;}cout << sum << endl;}}
0 0
- HDU 1397 Goldbach's Conjecture
- hdu 1397 Goldbach's Conjecture
- hdu 1397 Goldbach's Conjecture
- hdu 1397 Goldbach's Conjecture
- HDU - 1397 Goldbach's Conjecture
- HDU 1397 Goldbach's Conjecture
- 【HDU】 1397 Goldbach's Conjecture
- HDU 1397 Goldbach's Conjecture
- hdu 1397 Goldbach's Conjecture
- hdu Goldbach's Conjecture
- 1397 Goldbach's Conjecture
- HDOJ 1397 Goldbach's Conjecture
- HDU 1397 Goldbach's Conjecture(素数打表)
- 杭电 HDU ACM 1397 Goldbach's Conjecture
- POJ 2909 && HDU 1397 Goldbach's Conjecture(数论)
- HDU 1397 Goldbach's Conjecture(素数判断)
- HDU - 1397 Goldbach's Conjecture(哥德巴赫猜想)
- HDU 1397 Goldbach's Conjecture
- POJ - 1328 Radar Installation
- POJ - 3253 Fence Repair
- CodeForces - 484D Kindergarten
- HDU - 2028 Lowest Common Multiple Plus
- HDU - 1576 A/B
- HDU - 1397 Goldbach's Conjecture
- POJ - 3070 Fibonacci
- FZU - 2020 组合
- Linux中进程的几种状态
- CFLAGS=-static 静态编译 少林功夫真嘿侯
- HDU - 4497 GCD and LCM
- HDU - 3398 String
- POJ - 3624 Charm Bracelet
- HDU - 2546 饭卡