CodeForces 1038B

题目链接

https://vjudge.net/problem/CodeForces-1038B

题意

有1—n个数，问能否把他们分成两个非空集合S1,S2，使得 gcd(sum(S1),sum(S2)) > 1

思路

• 如果n=1或n=2，显然不行

• 如果n>2，通过观察可以发现，奇数分成一组，偶数分成一组，他们必有公因数2。

题解的解法更简单，因为n个数的和是 n*(n+1)/2，所以一个集合放n/2，一个集合放其他的，那么他们必有公因数k。

AC代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<string>
#include<sstream>
#include<cctype>
#include<map>
#include<stack>
#include<queue>
#include<list>
#include<cstdlib>
#include<ctime>
using namespace std;
typedef long long ll;
const double PI = atan(1.0)*4;
const int INF = 0x3f3f3f3f;
const ll MOD = 1000000007;
const int maxn = 100010;

int main()
{
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
    int n;
    scanf("%d", &n);
    if(n == 1 || n == 2)
        printf("No\n");
    else
    {
        printf("Yes\n");
        if(n % 2 == 0)
        {
            printf("%d", n / 2);
            for(int i = 2; i <= n; i = i + 2)
                printf(" %d", i);
            printf("\n%d", n / 2);
            for(int i = 1; i <= n; i = i + 2)
                printf(" %d", i);
            printf("\n");
        }
        else
        {
            printf("%d", n / 2);
            for(int i = 2; i <= n; i = i + 2)
                printf(" %d", i);
            printf("\n%d", n / 2 + 1);
            for(int i = 1; i <= n; i = i + 2)
                printf(" %d", i);
            printf("\n");
        }
    }
    return 0;
}