Codeforces Round #501 (Div. 3) 1015D Walking Between Houses

D. Walking Between Houses

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are $\mathrm{nn\; houses\; in\; a\; row.\; They\; are\; numbered\; from\; 11\; to\; nn\; in\; order\; from\; left\; to\; right.\; Initially\; you\; are\; in\; the\; house\; 11\; .}$

You have to perform $\mathrm{kk\; moves\; to\; other\; house.\; In\; one\; move\; you\; go\; from\; your\; current\; house\; to\; some\; other\; house.\; You\; can\text{'}t\; stay\; where\; you\; are\; (i.e.,\; in\; each\; move\; the\; new\; house\; differs\; from\; the\; current\; house).\; If\; you\; go\; from\; the\; house\; xx\; to\; the\; house\; yy\; ,\; the\; total\; distance\; you\; walked\; increases\; by\; |x-y||x-y|\; units\; of\; distance,\; where\; |a||a|\; is\; the\; absolute\; value\; of\; aa\; .\; It\; is\; possible\; to\; visit\; the\; same\; house\; multiple\; times\; (but\; you\; can\text{'}t\; visit\; the\; same\; house\; in\; sequence).}$

Your goal is to walk exactly $\mathrm{ss\; units\; of\; distance\; in\; total.}$

If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly $\mathrm{kk\; moves.}$

Input

The first line of the input contains three integers $\mathrm{nn\; ,\; kk\; ,\; ss\; (2\le n\le 1092\le n\le 109\; ,\; 1\le k\le 2\cdot 1051\le k\le 2\cdot 105\; ,\; 1\le s\le 10181\le s\le 1018\; )\; —\; the\; number\; of\; houses,\; the\; number\; of\; moves\; and\; the\; total\; distance\; you\; want\; to\; walk.}$

Output

If you cannot perform $\mathrm{kk\; moves\; with\; total\; walking\; distance\; equal\; to\; ss\; ,\; print\; "NO".}$

Otherwise print "YES" on the first line and then print exactly $\mathrm{kk\; integers\; hihi\; (1\le hi\le n1\le hi\le n\; )\; on\; the\; second\; line,\; where\; hihi\; is\; the\; house\; you\; visit\; on\; the\; ii\; -th\; move.}$

For each $\mathrm{jj\; from\; 11\; to\; k-1k-1\; the\; following\; condition\; should\; be\; satisfied:\; hj\ne hj+1hj\ne hj+1\; .\; Also\; h1\ne 1h1\ne 1\; should\; be\; satisfied.}$

Examples

Input

Copy

10 2 15

Output

Copy

YES
10 4 

Input

Copy

10 9 45

Output

Copy

YES
10 1 10 1 2 1 2 1 6 

Input

Copy

10 9 81

Output

Copy

YES
10 1 10 1 10 1 10 1 10 

Input

Copy

10 9 82

Output

Copy

NO

题目大意：1~n个房子，起点为1，然后规定刚好k步，走完s的距离，（从x到y，距离为|x-y|）.

思路：以为是个深搜。但是感觉写不了。。。看了官方题解。（蠢了）。左右走无法判断。

官方给了 cur + x 和 cur - x 来左右走。怎么保证刚好k步s距离呢。 s-(k-1)

如果s-(k-1) > (n-1)  走最大的距离，然后k -= 1,  s -= l;  (l为他俩的最小值)

当k等于1，剩最后s'，就保证刚好走完s。k != 1， 总还会剩一点。比较巧妙

不能刚好k步走完s，当且仅当 k > s or  k*(n-1) < s

能继续走，当且仅当 k-1 <=  s-x      x <= n-1 也就是min(n-1, s-(k-1));

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <sstream>
#include <stack>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;

ll step(ll cur, ll x) {
    if(cur - x > 0)//能左走就先左走 
        return cur - x;
    else//不行就右走
        return cur + x;
}

int main() {
    //freopen("in.txt", "r", stdin);
    ll n, k, s, cur = 1;
    scanf("%lld%lld%lld", &n, &k, &s);
    if(k > s || k*(n-1) < s) {//只有这两种情况NO 
        printf("NO
");
        return 0;
    }
    printf("YES
");
    while(k > 0) {//先走最大的，直到最后不足最大的，然后一点点走。 
        ll l = n-1 < s-(k-1) ? n-1 : s-(k-1);// s-(k-1)能保证k步刚好s距离 
        cur = step(cur, l);//左走还是右走 
        printf("%lld ", cur);
        s -= l;
        k -= 1;
    }
}