207 字
1 分钟
灌水
2017-08-12

20170812080054_73383.png

输入#

样例输入1:
3 1
样例输出1:
3 1 2
样例输入2:
4 1
样例输出2:
4 3 1 2
样例输入3:
8 17
样例输出3:
6 2 3 1 8 4 5 7

提示

20170812080119_26860.png

题解#

先找规律然后瞎搞就可以了

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define LL long long
LL Sum[1000005];
LL x;
int ans[1000005];
int main()
{
    int n;
    scanf("%d%lld", &n, &x);
    for (int i = n - 2; i >= 0; i--)
        Sum[i] = Sum[i + 1] + i;
    if (Sum[0] < x)
    {
        puts("-1");
        return 0;
    }
    int k, i, j;
    for (k = 1; k <= n - 2; k++)
        if (x > Sum[k])
            break;
    x = x - Sum[k];
    for (i = n - k, j = 1; i <= n - 1; i++, j++)
        ans[j] = i;
    for (i = 1; i <= n - k - 1; i++, j++)
        ans[j] = i;
    ans[n] = n;
    i = 1;
    while (x--)
    {
        swap(ans[i], ans[i + 1]);
        i++;
    }
    for (i = 1; i < n; i++)
        printf("%d ", ans[i]);
    printf("%d\n", ans[n]);
}
灌水
https://www.nekomio.com/posts/88/
作者
NekoMio
发布于
2017-08-12
许可协议
CC BY-NC-SA 4.0