Codeforces Round #486 (Div. 3) CD

C. Equal Sums

You are given $\mathrm{kk sequences\; of\; integers.\; The\; length\; of\; the ii-th\; sequence\; equals\; to nini.}$

You have to choose exactly two sequences $\mathrm{ii and jj (i\ne ji\ne j)\; such\; that\; you\; can\; remove\; exactly\; one\; element\; in\; each\; of\; them\; in\; such\; a\; way\; that\; the\; sum\; of\; the\; changed\; sequence ii (its\; length\; will\; be\; equal\; to ni-1ni-1)\; equals\; to\; the\; sum\; of\; the\; changed\; sequence jj (its\; length\; will\; be\; equal\; to nj-1nj-1).}$

Note that it's required to remove exactly one element in each of the two chosen sequences.

Assume that the sum of the empty (of the length equals $\mathrm{00)\; sequence\; is 00.}$

Input

The first line contains an integer $\mathrm{kk (2\le k\le 2\cdot 1052\le k\le 2\cdot 105)\; —\; the\; number\; of\; sequences.}$

Then $\mathrm{kk pairs\; of\; lines\; follow,\; each\; pair\; containing\; a\; sequence.}$

The first line in the $\mathrm{ii-th\; pair\; contains\; one\; integer nini (1\le ni<2\cdot 1051\le ni<2\cdot 105)\; —\; the\; length\; of\; the ii-th\; sequence.\; The\; second\; line\; of\; the ii-th\; pair\; contains\; a\; sequence\; of nini integers ai,1,ai,2,\dots ,ai,niai,1,ai,2,\dots ,ai,ni.}$

The elements of sequences are integer numbers from $-104-104 to 104104.$

The sum of lengths of all given sequences don't exceed $\mathrm{2\cdot 1052\cdot 105,\; i.e. n1+n2+\cdots +nk\le 2\cdot 105n1+n2+\cdots +nk\le 2\cdot 105.}$

Output

If it is impossible to choose two sequences such that they satisfy given conditions, print "NO" (without quotes). Otherwise in the first line print "YES" (without quotes), in the second line — two integers $\mathrm{ii, xx (1\le i\le k,1\le x\le ni1\le i\le k,1\le x\le ni),\; in\; the\; third\; line\; —\; two\; integers jj, yy (1\le j\le k,1\le y\le nj1\le j\le k,1\le y\le nj).\; It\; means\; that\; the\; sum\; of\; the\; elements\; of\; the ii-th\; sequence\; without\; the\; element\; with\; index xx equals\; to\; the\; sum\; of\; the\; elements\; of\; the jj-th\; sequence\; without\; the\; element\; with\; index yy.}$

Two chosen sequences must be distinct, i.e. $\mathrm{i\ne ji\ne j.\; You\; can\; print\; them\; in\; any\; order.}$

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If there are multiple possible answers, print any of them.

Examples

input

Copy

2
5
2 3 1 3 2
6
1 1 2 2 2 1

output

Copy

YES
2 6
1 2

input

Copy

3
1
5
5
1 1 1 1 1
2
2 3

output

Copy

NO

input

Copy

4
6
2 2 2 2 2 2
5
2 2 2 2 2
3
2 2 2
5
2 2 2 2 2

output

Copy

YES
2 2
4 1

Note

In the first example there are two sequences $[2,3,1,3,2][2,3,1,3,2] and [1,1,2,2,2,1][1,1,2,2,2,1].\; You\; can\; remove\; the\; second\; element\; from\; the\; first\; sequence\; to\; get [2,1,3,2][2,1,3,2] and\; you\; can\; remove\; the\; sixth\; element\; from\; the\; second\; sequence\; to\; get [1,1,2,2,2][1,1,2,2,2].\; The\; sums\; of\; the\; both\; resulting\; sequences\; equal\; to 88,\; i.e.\; the\; sums\; are\; equal.$

任选两个序列，两个序列都除去他们中的一个数，使的总和相同。

因为总和的数量不超过2e5   所以能全部用结构体保存。 有三个值。

假设第i个序列，第j个数。  那么sum是第i个序列总和减去第j个数的值，id是第几个数。kk是第几个序列。

那么排下序，找到相同的sum，然后在找不同的kk  存在就是YES 或者就是 NO。

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 const int N = 2e5+10;
 5 ll k, len, x;
 6 struct Nod{
 7     ll sum, id, kk;
 8     Nod() {
 9         id = kk = sum = -1;
10     }
11 }nod[N];
12 ll a[N];
13 bool cmp(const Nod &a, const Nod &b) {
14     if(a.sum != b.sum ) return a.sum < b.sum;
15     else return a.kk < b.kk;
16 }
17 int main() {
18     int n = 0;
19     scanf("%lld", &k);
20     for(int i = 1; i <= k; i ++) {
21         scanf("%lld", &len);
22         ll sum = 0;
23         for(int j = 1; j <= len; j ++) {
24             scanf("%lld", &a[j]);
25             sum += a[j];
26         }
27         for(int j = 1; j <= len; j ++) {
28             nod[n].sum = sum - a[j];
29             nod[n].kk = i;
30             nod[n++].id = j;
31         }
32     }
33     sort(nod,nod+n,cmp);
34     // for(int i = 0; i < n; i ++) {
35     //     printf("%lld %lld %lld\n",nod[i].sum,nod[i].kk,nod[i].id);
36     // }
37     int l = 0, r = 1;
38     while(r < n) {
39         while(nod[l].sum != nod[r].sum && r+1 < n) {
40             l++;r++;
41         }
42         while(nod[l].sum == nod[r].sum && nod[r].sum == nod[r+1].sum) r++;
43         if(nod[l].sum == nod[r].sum && nod[l].kk != nod[r].kk) {
44             printf("YES\n");
45             return 0*printf("%lld %lld\n%lld %lld\n",nod[l].kk,nod[l].id,nod[r].kk,nod[r].id);
46         }
47         r += 2;
48         l = r - 1;
49     }
50     printf("NO\n");
51     return 0;
52 }

D. Points and Powers of Two

standard output

There are $\mathrm{nn distinct\; points\; on\; a\; coordinate\; line,\; the\; coordinate\; of ii-th\; point\; equals\; to xixi.\; Choose\; a\; subset\; of\; the\; given\; set\; of\; points\; such\; that\; the\; distance\; between\; each\; pair\; of\; points\; in\; a\; subset\; is\; an\; integral\; power\; of\; two.\; It\; is\; necessary\; to\; consider\; each\; pair\; of\; points,\; not\; only\; adjacent.\; Note\; that\; any\; subset\; containing\; one\; element\; satisfies\; the\; condition\; above.\; Among\; all\; these\; subsets,\; choose\; a\; subset\; with\; maximum\; possible\; size.}$

In other words, you have to choose the maximum possible number of points ${\mathrm{xi1,xi2,\dots ,ximxi1,xi2,\dots ,xim such\; that\; for\; each\; pair xijxij, xikxik it\; is\; true\; that |xij-xik|=2d|xij-xik|=2d where dd is\; some\; non-negative\; integer\; number\; (not\; necessarily\; the\; same\; for\; each\; pair\; of\; points).}}^{}$

Input

The first line contains one integer $\mathrm{nn (1\le n\le 2\cdot 1051\le n\le 2\cdot 105)\; —\; the\; number\; of\; points.}$

The second line contains $\mathrm{nn pairwise\; distinct\; integers x1,x2,\dots ,xnx1,x2,\dots ,xn (-109\le xi\le 109-109\le xi\le 109)\; —\; the\; coordinates\; of\; points.}$

Output

In the first line print $\mathrm{mm —\; the\; maximum\; possible\; number\; of\; points\; in\; a\; subset\; that\; satisfies\; the\; conditions\; described\; above.}$

In the second line print $\mathrm{mm integers\; —\; the\; coordinates\; of\; points\; in\; the\; subset\; you\; have\; chosen.}$

If there are multiple answers, print any of them.

Examples

input

Copy

6
3 5 4 7 10 12

output

Copy

3
7 3 5

input

Copy

5
-1 2 5 8 11

output

Copy

1
8

Note

In the first example the answer is $[7,3,5][7,3,5].\; Note,\; that |7-3|=4=22|7-3|=4=22, |7-5|=2=21|7-5|=2=21 and |3-5|=2=21|3-5|=2=21.\; You\; can\text{'}t\; find\; a\; subset\; having\; more\; points\; satisfying\; the\; required\; property.$

 选取最大的区间，使的两两之差是2^d ，d是非负整数。可以得出一个结论，最大的子集不会超过3个。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int N = 2e5+10;
 4 int a[N], n;
 5 map<int,int> mp;
 6 int main() {
 7     scanf("%d",&n);
 8     for(int i = 0; i < n; i ++) {
 9         scanf("%d",&a[i]);
10         mp[a[i]] = 1;
11     }
12     sort(a,a+n);
13     for(int i = 0; i < n; i ++) {
14         for(int j = 0; j < 32; j ++) {
15             int x = 1 << j;
16             if(mp[a[i]+x] && mp[a[i]+2*x]) {
17                 return 0*printf("3\n%d %d %d\n",a[i],a[i]+x,a[i]+x*2);
18             }
19             if(a[i]+x > a[n-1] || a[i]+2*x > a[n-1]) break;
20         }
21     }
22     for(int i = 0; i < n; i ++) {
23         for(int j = 0; j < 32; j ++) {
24             int x = 1 << j;
25             if(mp[a[i]+x]) {
26                 return 0*printf("2\n%d %d\n",a[i],a[i]+x);
27             }
28             if(mp[a[i]+x*2]) {
29                 return 0*printf("2\n%d %d\n",a[i],a[i]+2*x);
30             }
31             if(a[i]+x > a[n-1] || a[i]+2*x > a[n-1]) break;
32         }
33     }
34     printf("1\n%d\n",a[0]);
35     return 0;
36 }