Walking Between Houses（贪心+思维）

Walking Between Houses

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

10 2 15

output

YES
10 4 

input

10 9 45

output

YES
10 1 10 1 2 1 2 1 6 

input

10 9 81

output

YES
10 1 10 1 10 1 10 1 10 

input

10 9 82

output

NO


题目意思：

现在有n个房子排成一列，编号为1~n，起初你在第1个房子里，现在你要进行k次移动，每次移动一都可以从一个房子i移动到另外一个其他的房子j
里（i != j），移动的距离为|j - i|。问你进过k次移动后，移动的总和可以刚好是s吗？若可以则输出YES并依次输出每次到达的房子的编号，
否则输出NO。

解题思路：
每次至少移动一个单位的距离，至多移动n-1个单位的距离，所以要想完成上述要求每次决策前后一定要满足条件： 
k <= s && k*(n-1) >= s

   假设当前决策为移动x单位的距离，所以要满足下述条件： 
 （ k-1 <= s-x） && （x <= n-1）

   得：max(x) = min( (s - k + 1) , (n-1) ) 
   因此我们的决策为：每次移动的大小为 s与k的差值 和 n-1 中的较小值，使得s和k尽快的相等。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<cstring>
 4 #define ll long long int
 5 #include<algorithm>
 6 using namespace std;
 7 ll n,s,k;
 8 int main()
 9 {
10     ll sum,i,pos,len;
11     scanf("%lld%lld%lld",&n,&k,&s);
12     if(k>s||k*(n-1)<s)
13     {
14         printf("NO\n");
15     }
16     else
17     {
18         printf("YES\n");
19         pos=1;///记录位置
20         while(k--)
21         {
22           len=min(s-k,n-1);
23           s=s-len;
24           if(pos+len<=n)///向前移还是向后移的判断
25           {
26               pos=pos+len;
27           }
28           else
29           {
30               pos=pos-len;
31           }
32           printf("%d ",pos);
33         }
34     }
35     return 0;
36 }