SDNUOJ-1500-哥德巴赫猜想-素数-二分

Description
 Spring is coming, T_T.
 All you know Goldbach conjecture.That is to say, Every even integer greater than 2 can be expressed as the sum of two primes. Today, skywind present a new conjecture: every even integer can be expressed as the difference of two primes. To validate this conjecture, you are asked to write a program.

Input
 The first line of input is a number n identified the count of test cases(n<10^4). There is a even number x at the next n lines. The absolute value of x is not greater than 10^6.
Output
 For each number x tested, outputs two primes a and b at one line separated with one space where a-b=x. If more than one group can meet it, output the minimum group(of course refers b). If no primes can satisfy it, output 'FAIL'.

Sample Input
 3
 6
 10
 20
Sample Output
 11 5
 13 3
 23 3

题意:验证哥德巴赫猜想,任一大于2的偶数都可写成两个质数之和。给出一个数N,求满足N = 素数A – 素数B的结果。结果会有很多,因此要求B最小的结果。

思路:
N = 素数A – 素数B
移项,可得
N + 素数B = 素数A(求B最小的情况)
因此可以利用素数筛打表,然后从最小的素数跑暴力。

即:判断 N + 素数B 是否为素数,可以在之前打的素数表中进行二分查找,如果能找到,则说明结果正确。

#include <iostream>
#include <cstdio>

using namespace std;
const int TOP = 1e6 + 100000;
bool e[TOP];
int p[TOP / 5];
int pnum;

void init() {
    e[0] = e[1] = 1;
    pnum = 0;
    for (int i = 2; i < TOP; i++) {
        if (e[i] == 0)p[++pnum] = i;
        for (int j = 1; j <= pnum && p[j] * i < TOP; j++) {
            e[p[j] * i] = 1;
            if (i % p[j] == 0)break;
        }
    }
}

bool binSearch(int n) {
    int l = 1;
    int r = pnum - 1;
    int mid = (l + r) / 2;
    while (l != mid) {
        mid = (l + r) / 2;
        if (p[mid] == n) {
            return true;
        } else if (p[mid] > n) {
            r = mid;
        } else {
            l = mid;
        }
        mid = (l + r) / 2;
    }

    return false;
}

int main() {
    init();
    int T, tmp;
    scanf("%d", &T);
    while (T--) {
        scanf("%d", &tmp);
        if (tmp == 0) {
            printf("2 2\n");
            continue;
        }
        for (int i = 1; i < pnum; i++) {
            if (binSearch(p[i] + tmp)) {
                printf("%d %d\n", p[i] + tmp, p[i]);
                break;
            }
        }
    }
    return 0;
}

Azure99

大二蒟蒻,喜欢折腾vps、玩机,偶尔写写代码

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