Codeforces Round #479 (Div. 3) ABCDEF

A. Wrong Subtraction

Little girl Tanya is learning how to decrease a number by one, but she does it wrong with a number consisting of two or more digits. Tanya subtracts one from a number by the following algorithm:

• if the last digit of the number is non-zero, she decreases the number by one;
• if the last digit of the number is zero, she divides the number by 10 (i.e. removes the last digit).

You are given an integer number $\mathrm{nn.\; Tanya\; will\; subtract\; one\; from\; it kk times.\; Your\; task\; is\; to\; print\; the\; result\; after\; all kk subtractions.}$

It is guaranteed that the result will be positive integer number.

Input

The first line of the input contains two integer numbers $\mathrm{nn and kk (2\le n\le 1092\le n\le 109, 1\le k\le 501\le k\le 50)\; —\; the\; number\; from\; which\; Tanya\; will\; subtract\; and\; the\; number\; of\; subtractions\; correspondingly.}$

Output

Print one integer number — the result of the decreasing $\mathrm{nn by\; one kk times.}$

It is guaranteed that the result will be positive integer number.

Examples

input

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512 4

output

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50

input

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1000000000 9

output

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1

Note

The first example corresponds to the following sequence: $\mathrm{512\to 511\to 510\to 51\to 50512\to 511\to 510\to 51\to 50.}$

 

$\mathrm{签到}$

 1 #include 
 2 #define ll long long
 3 using namespace std;
 4 int main() {
 5     int n, k;
 6     cin >> n >> k;
 7     for(int i = 0; i < k; i ++) {
 8         if(n%10) n--;
 9         else n/=10;
10     }
11     cout << n << endl;
12     return 0;
13 }

B. Two-gram

Two-gram is an ordered pair (i.e. string of length two) of capital Latin letters. For example, "AZ", "AA", "ZA" — three distinct two-grams.

You are given a string $\mathrm{ss consisting\; of nn capital\; Latin\; letters.\; Your\; task\; is\; to\; find any two-gram\; contained\; in\; the\; given\; string as\; a\; substring(i.e.\; two\; consecutive\; characters\; of\; the\; string)\; maximal\; number\; of\; times.\; For\; example,\; for\; string ss =\; "BBAABBBA"\; the\; answer\; is\; two-gram\; "BB",\; which\; contained\; in ss three\; times.\; In\; other\; words,\; find\; any\; most\; frequent\; two-gram.}$

Note that occurrences of the two-gram can overlap with each other.

Input

The first line of the input contains integer number $\mathrm{nn (2\le n\le 1002\le n\le 100)\; —\; the\; length\; of\; string ss.\; The\; second\; line\; of\; the\; input\; contains\; the\; string ss consisting\; of nn capital\; Latin\; letters.}$

Output

Print the only line containing exactly two capital Latin letters — any two-gram contained in the given string $\mathrm{ss as\; a\; substring (i.e.\; two\; consecutive\; characters\; of\; the\; string)\; maximal\; number\; of\; times.}$

Examples

input

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7
ABACABA

output

Copy

AB

input

Copy

5
ZZZAA

output

Copy

ZZ

Note

In the first example "BA" is also valid answer.

In the second example the only two-gram "ZZ" can be printed because it contained in the string "ZZZAA" two times.

求出现最多的连续两个字母

 1 #include 
 2 #define ll long long
 3 using namespace std;
 4 map<string,int> mp;
 5 int main() {
 6     int n;
 7     string s;
 8     cin >> n >> s;
 9     for(int i = 0; i < n-1; i ++) {
10         string ss = s.substr(i,2);
11         mp[ss]++;
12     }
13     n = 0;
14     map<string,int> ::iterator it = mp.begin();
15     for(; it!=mp.end(); it++) {
16         if((*it).second > n) {
17             n = (*it).second;
18             s = (*it).first;
19         }
20     }
21     cout << s<< endl;
22     return 0;
23 }

C. Less or Equal

You are given a sequence of integers of length $\mathrm{nn and\; integer\; number kk.\; You\; should\; print any\; integer number xx in\; the\; range\; of [1;109][1;109] (i.e. 1\le x\le 1091\le x\le 109)\; such\; that\; exactly kk elements\; of\; given\; sequence\; are\; less\; than\; or\; equal\; to xx.}$

Note that the sequence can contain equal elements.

If there is no such $\mathrm{xx,\; print\; "-1"\; (without\; quotes).}$

Input

The first line of the input contains integer numbers $\mathrm{nn and kk (1\le n\le 2\cdot 1051\le n\le 2\cdot 105, 0\le k\le n0\le k\le n).\; The\; second\; line\; of\; the\; input\; contains nn integer\; numbers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 1091\le ai\le 109)\; —\; the\; sequence\; itself.}$

Output

Print any integer number $\mathrm{xx from\; range [1;109][1;109] such\; that\; exactly kk elements\; of\; given\; sequence\; is\; less\; or\; equal\; to xx.}$

If there is no such $\mathrm{xx,\; print\; "-1"\; (without\; quotes).}$

Examples

input

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7 4
3 7 5 1 10 3 20

output

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6

input

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7 2
3 7 5 1 10 3 20

output

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-1

Note

In the first example $\mathrm{55 is\; also\; a\; valid\; answer\; because\; the\; elements\; with\; indices [1,3,4,6][1,3,4,6] is\; less\; than\; or\; equal\; to 55 and\; obviously\; less\; than\; or\; equal\; to 66.}$

In the second example you cannot choose any number that only $\mathrm{22 elements\; of\; the\; given\; sequence\; will\; be\; less\; than\; or\; equal\; to\; this\; number\; because 33 elements\; of\; the\; given\; sequence\; will\; be\; also\; less\; than\; or\; equal\; to\; this\; number.}$

 1 #include 
 2 #define ll long long
 3 using namespace std;
 4 const int N = 2e5+10;
 5 int a[N];
 6 int main() {
 7     int n, k;
 8     cin >> n >> k;
 9     for(int i = 0; i < n; i ++) cin >> a[i];
10     sort(a,a+n);
11     if(k == 0) {
12         if(a[0] == 1) printf("-1\n");
13         else printf("1\n");
14         return 0;
15     }
16     if(a[k-1] == a[k]) printf("-1\n");
17     else printf("%d\n",a[k-1]);
18     return 0;
19 }

D. Divide by three, multiply by two

Polycarp likes to play with numbers. He takes some integer number $\mathrm{xx,\; writes\; it\; down\; on\; the\; board,\; and\; then\; performs\; with\; it n-1n-1operations\; of\; the\; two\; kinds:}$

• divide the number $\mathrm{xx by 33 (xx must\; be\; divisible\; by 33);}$
• multiply the number $\mathrm{xx by 22.}$

After each operation, Polycarp writes down the result on the board and replaces $\mathrm{xx by\; the\; result.\; So\; there\; will\; be nn numbers\; on\; the\; board\; after\; all.}$

You are given a sequence of length $\mathrm{nn —\; the\; numbers\; that\; Polycarp\; wrote\; down.\; This\; sequence\; is\; given\; in\; arbitrary\; order,\; i.e.\; the\; order\; of\; the\; sequence\; can\; mismatch\; the\; order\; of\; the\; numbers\; written\; on\; the\; board.}$

Your problem is to rearrange (reorder) elements of this sequence in such a way that it can match possible Polycarp's game in the order of the numbers written on the board. I.e. each next number will be exactly two times of the previous number or exactly one third of previous number.

It is guaranteed that the answer exists.

Input

The first line of the input contatins an integer number $\mathrm{nn (2\le n\le 1002\le n\le 100)\; —\; the\; number\; of\; the\; elements\; in\; the\; sequence.\; The\; second\; line\; of\; the\; input\; contains nn integer\; numbers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 3\cdot 10181\le ai\le 3\cdot 1018)\; —\; rearranged\; (reordered)\; sequence\; that\; Polycarp\; can\; wrote\; down\; on\; the\; board.}$

Output

Print $\mathrm{nn integer\; numbers\; —\; rearranged\; (reordered)\; input\; sequence\; that\; can\; be\; the\; sequence\; that\; Polycarp\; could\; write\; down\; on\; the\; board.}$

It is guaranteed that the answer exists.

Examples

input

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6
4 8 6 3 12 9

output

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9 3 6 12 4 8 

input

Copy

4
42 28 84 126

output

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126 42 84 28 

input

Copy

2
1000000000000000000 3000000000000000000

output

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3000000000000000000 1000000000000000000 

Note

In the first example the given sequence can be rearranged in the following way: $[9,3,6,12,4,8][9,3,6,12,4,8].\; It\; can\; match\; possible\; Polycarp\text{'}s\; game\; which\; started\; with x=9x=9.$

pre[i] 表示第i个是的前面是第j个数，nex[i] 表示第i个数的后面是第j个数

 1 #include 
 2 #define ll long long
 3 using namespace std;
 4 ll a[110], nex[110], pre[110];
 5 int main() {
 6     int n;
 7     cin >> n;
 8     for(int i = 1; i <= n; i ++) nex[i] = pre[i] = i;
 9     for(int i = 1; i <= n; i ++) cin >> a[i];
10     for(int i = 1; i <= n; i ++) {
11         for(int j = 1; j <= n; j ++) {
12             if(a[i] > a[j]) {
13                 if(a[i]%3==0 && a[i]/3 == a[j]) {
14                     nex[i] = j;
15                     pre[j] = i;
16                 } else if(a[i]%2==0&&a[i]/2==a[j]) {
17                     nex[j] = i;
18                     pre[i] = j;
19                 }
20             } else if(a[j] > a[i]){
21                 if(a[j]%3==0 && a[j]/3 == a[i]) {
22                     nex[j] = i;
23                     pre[i] = j;
24                 } else if(a[j]%2==0&&a[j]/2==a[i]) {
25                     nex[i] = j;
26                     pre[j] = i;
27                 }
28             }
29         }
30     }
31     int id = 0;
32     for(int i = 1; i <= n; i ++) {
33         if(pre[i] == i) {
34             id = i;
35             break;
36         }
37     }
38     while(nex[id] != id) {
39         printf("%lld ",a[id]);
40         id = nex[id];
41     }
42     printf("%lld\n",a[id]);
43     return 0;
44 }

E. Cyclic Components

You are given an undirected graph consisting of $\mathrm{nn vertices\; and mm edges.\; Your\; task\; is\; to\; find\; the\; number\; of\; connected\; components\; which\; are\; cycles.}$

Here are some definitions of graph theory.

An undirected graph consists of two sets: set of nodes (called vertices) and set of edges. Each edge connects a pair of vertices. All edges are bidirectional (i.e. if a vertex $\mathrm{aa is\; connected\; with\; a\; vertex bb,\; a\; vertex bb is\; also\; connected\; with\; a\; vertex aa).\; An\; edge\; can\text{'}t\; connect\; vertex\; with\; itself,\; there\; is\; at\; most\; one\; edge\; between\; a\; pair\; of\; vertices.}$

Two vertices $\mathrm{uu and vv belong\; to\; the\; same\; connected\; component\; if\; and\; only\; if\; there\; is\; at\; least\; one\; path\; along\; edges\; connecting uu and vv.}$

A connected component is a cycle if and only if its vertices can be reordered in such a way that:

• the first vertex is connected with the second vertex by an edge,
• the second vertex is connected with the third vertex by an edge,
• ...
• the last vertex is connected with the first vertex by an edge,
• all the described edges of a cycle are distinct.

A cycle doesn't contain any other edges except described above. By definition any cycle contains three or more vertices.

There are $\mathrm{66 connected\; components, 22 of\; them\; are\; cycles: [7,10,16][7,10,16] and [5,11,9,15][5,11,9,15].}$

Input

The first line contains two integer numbers $\mathrm{nn and mm (1\le n\le 2\cdot 1051\le n\le 2\cdot 105, 0\le m\le 2\cdot 1050\le m\le 2\cdot 105)\; —\; number\; of\; vertices\; and\; edges.}$

The following $\mathrm{mm lines\; contains\; edges:\; edge ii is\; given\; as\; a\; pair\; of\; vertices vivi, uiui (1\le vi,ui\le n1\le vi,ui\le n, ui\ne viui\ne vi).\; There\; is\; no\; multiple\; edges\; in\; the\; given\; graph,\; i.e.\; for\; each\; pair\; (vi,uivi,ui)\; there\; no\; other\; pairs\; (vi,uivi,ui)\; and\; (ui,viui,vi)\; in\; the\; list\; of\; edges.}$

Output

Print one integer — the number of connected components which are also cycles.

Examples

input

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5 4
1 2
3 4
5 4
3 5

output

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1

input

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17 15
1 8
1 12
5 11
11 9
9 15
15 5
4 13
3 13
4 3
10 16
7 10
16 7
14 3
14 4
17 6

output

Copy

2

Note

In the first example only component $[3,4,5][3,4,5] is\; also\; a\; cycle.$

The illustration above corresponds to the second example.

判断有多少个圈。构造圈的话，圈中的每个节点都有两个节点，dfs遍历。

 1 #include 
 2 #define ll long long
 3 using namespace std;
 4 const int N = 2e5+10;
 5 vector<int> vs[N];
 6 bool vis[N], flag;
 7 int root;
 8 void dfs(int x, int f, int de) {
 9     vis[x] = 1;
10     if(vs[x].size() != 2) return;
11     for(int i = 0; i < vs[x].size(); i ++) {
12         int v = vs[x][i];
13         if(!vis[v] && v != f) dfs(v, x,de+1);
14         if(v == root && de != 1) {
15             flag = true;
16             return;
17         }
18     }
19 }
20 int main() {
21     int n, m;
22     cin >> n >> m;
23     for(int i = 0; i < m; i ++) {
24         int u, v;
25         cin >> u >> v;
26         vs[v].push_back(u);
27         vs[u].push_back(v);
28     }
29     int ans = 0;
30     for(int i = 1; i <= n; i ++) {
31         if(!vis[i]) {
32             flag = 0;
33             root = i;
34             dfs(i,-1,0);
35             if(flag) ans++;
36         }
37     }
38     cout << ans << endl;
39     return 0;
40 }

F. Consecutive Subsequence

You are given an integer array of length $\mathrm{nn.}$

You have to choose some subsequence of this array of maximum length such that this subsequence forms a increasing sequence of consecutive integers. In other words the required sequence should be equal to $[x,x+1,\dots ,x+k-1][x,x+1,\dots ,x+k-1] for\; some\; value xx and\; length kk.$

Subsequence of an array can be obtained by erasing some (possibly zero) elements from the array. You can erase any elements, not necessarily going successively. The remaining elements preserve their order. For example, for the array $[5,3,1,2,4][5,3,1,2,4] the\; following\; arrays\; are\; subsequences: [3][3], [5,3,1,2,4][5,3,1,2,4], [5,1,4][5,1,4],\; but\; the\; array [1,3][1,3] is\; not.$

Input

The first line of the input containing integer number $\mathrm{nn (1\le n\le 2\cdot 1051\le n\le 2\cdot 105)\; —\; the\; length\; of\; the\; array.\; The\; second\; line\; of\; the\; input\; containing nn integer\; numbers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 1091\le ai\le 109)\; —\; the\; array\; itself.}$

Output

On the first line print $\mathrm{kk —\; the\; maximum\; length\; of\; the\; subsequence\; of\; the\; given\; array\; that\; forms\; an\; increasing\; sequence\; of\; consecutive\; integers.}$

On the second line print the sequence of the indices of the any maximum length subsequence of the given array that forms an increasing sequence of consecutive integers.

Examples

input

Copy

7
3 3 4 7 5 6 8

output

Copy

4
2 3 5 6 

input

Copy

6
1 3 5 2 4 6

output

Copy

2
1 4 

input

Copy

4
10 9 8 7

output

Copy

1
1 

input

Copy

9
6 7 8 3 4 5 9 10 11

output

Copy

6
1 2 3 7 8 9 

Note

All valid answers for the first example (as sequences of indices):

• $[1,3,5,6][1,3,5,6]$
• $[2,3,5,6][2,3,5,6]$

All valid answers for the second example:

• $[1,4][1,4]$
• $[2,5][2,5]$
• $[3,6][3,6]$

All valid answers for the third example:

• $[1][1]$
• $[2][2]$
• $[3][3]$
• $[4][4]$

All valid answers for the fourth example:

• $[1,2,3,7,8,9][1,2,3,7,8,9]$

•  

 判断最大的单调递增子序列，只能增1.

每输入一个数x，就判断x-1在之前是否存在，存在的话使用x的最长增系列长度是mp[x-1] +1 ，否则让mp[x] = 1。

这样mp[a[i]] 的最大值就是最长增序列了，然后输出下标就行。

 1 #include 
 2 #define INF 0x3f3f3f3f
 3 #define ll long long
 4 using namespace std;
 5 const int N = 2e5+10;
 6 map<int,int> mp;
 7 int a[N];
 8 int main() {
 9     int n, x;
10     cin >> n;
11     for(int i = 1; i <= n; i ++) {
12         cin >> a[i];
13         if(mp[a[i]-1]) {
14             mp[a[i]] = mp[a[i]-1]+1;
15         } else mp[a[i]] = 1;
16     }
17     map<int,int> ::iterator it = mp.begin();
18     int MAX = 0, id;
19     for(;it != mp.end(); ++it) {
20         if(MAX < (*it).second){
21             MAX = (*it).second;
22             id = (*it).first;
23         }
24     }
25     cout << MAX << endl;
26     int MIN = id - MAX + 1;
27     for(int i = 1; i <= n; i ++) {
28         if(a[i] == MIN) {
29             printf("%d ",i);
30             MIN++;
31         }
32         if(MIN > id) break;
33     }
34     return 0;
35 }

 

转载于:https://www.cnblogs.com/xingkongyihao/p/9028373.html