Codeforces 584D Dima and Lisa 【数学】

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本文介绍了一种解决特定问题的方法，即如何将一个给定的奇数表示为最多三个素数的和。通过使用暴力构造法，文章提供了一个简单且有效的解决方案，并附带了AC代码。

题目链接：Codeforces 584D Dima and Lisa

D. Dima and Lisa
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Dima loves representing an odd number as the sum of multiple primes, and Lisa loves it when there are at most three primes. Help them to represent the given number as the sum of at most than three primes.

More formally, you are given an odd numer n. Find a set of numbers pi (1 ≤ i ≤ k), such that

1 ≤ k ≤ 3
pi is a prime

The numbers pi do not necessarily have to be distinct. It is guaranteed that at least one possible solution exists.

Input
The single line contains an odd number n (3 ≤ n < 109).

Output
In the first line print k (1 ≤ k ≤ 3), showing how many numbers are in the representation you found.

In the second line print numbers pi in any order. If there are multiple possible solutions, you can print any of them.

Examples
input
27
output
3
5 11 11
Note
A prime is an integer strictly larger than one that is divisible only by one and by itself.

题意：构造不超过三个素数使得它们和为n。

思路：暴力构造，10^9内两个素数距离最大不会超过300。n = n - a + a - b + b，然后纯暴力就好了。方法不唯一。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#include <cmath>
#define fi first
#define se second
#define ll o<<1
#define rr o<<1|1
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MOD = 1e4 + 7;
const int MAXN = 1e8;
void add(LL &x, LL y) { x += y; x %= MOD; }
bool judge(int x) {
    if(x == 1) return false;
    for(int i = 2; i * i <= x; i++) {
        if(x % i == 0) {
            return false;
        }
    }
    return true;
}
int main()
{
    int n;
    while(scanf("%d", &n) != EOF) {
        if(judge(n)) {
            printf("1\n%d\n", n);
        }
        else if(judge(n-2)) {
            printf("2\n%d %d\n", 2, n-2);
        }
        else {
            int a, b, c;
            for(int i = 4; ; i++) {
                if(judge(n-i)) {
                    a = i; break;
                }
            }
            for(int i = 2; ; i++) {
                if(judge(a-i) && judge(i)) {
                    b = i; break;
                }
            }
            printf("3\n%d %d %d\n", n-a, a-b, b);
        }
    }
    return 0;
}