Codeforces Round #479 (Div. 3)题解

CF首次推出div3给我这种辣鸡做，当然得写份博客纪念下

A. Wrong Subtraction

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

Copy

512 4

output

Copy

50

input

Copy

1000000000 9

output

Copy

1

Note

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

分析：水题，就是末尾是0就除10，不为0就-1

#include <bits/stdc++.h>
using namespace std;

#define FF(i, a, b) for(int i = a; i < b; i++)
#define RR(i, a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define PR(x) printf("%d\n", x)
#define INF 0x3f3f3f3f
#define MAX 1001
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP; 

int main(void){
    #ifdef LOCAL
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
    #endif
    ios::sync_with_stdio(false); cin.tie(0);
    int i, n, k;
    
    cin>>n>>k;
    for(i = 0; i < k; i++){
        if(n % 10 != 0) n -= 1;
        else n /= 10;
    }
    cout<<n<<endl;
    return EXIT_SUCCESS;
}

B. Two-gram

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

Copy

7
ABACABA

output

Copy

AB

input

Copy

5
ZZZAA

output

Copy

ZZ

分析：水题，二维哈希映射可以做，我是直接string.find()查的

#include <bits/stdc++.h>
using namespace std;

#define FF(i, a, b) for(int i = a; i < b; i++)
#define RR(i, a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define PR(x) printf("%d\n", x)
#define INF 0x3f3f3f3f
#define MAX 1001
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP; 

int main(void){
    #ifdef LOCAL
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
    #endif
    ios::sync_with_stdio(false); cin.tie(0);
    int i, m, len, loca = -1 , max = 0, num[MAX] = {0}, flag; string str, str1;
    
    cin>>len>>str;
    for(i = 0; i < len - 1; i++){ 
        str1 = "";m = 0;str1 += str[i]; str1 += str[i + 1];
        flag = str.find(str1, 0);
        while(flag != string::npos){num[i]++;flag = str.find(str1, flag + 1);}
    }
    for(i = 0; i < len - 1; i++)if(num[i] > max){loca = i;max = num[i];}

    cout<<str[loca]<<str[loca + 1]<<endl; 
    
    return EXIT_SUCCESS;
}

C. Less or Equal

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

Copy

7 4
3 7 5 1 10 3 20

output

Copy

6

input

Copy

7 2
3 7 5 1 10 3 20

output

Copy

-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 $\mathrm{22 elements\; of\; the\; given\; sequence\; will\; be\; less\; than\; or\; equal\; to\; this\; number\; because 33 elements\; of\; the\; given\; sequence\; will\; be\; also\; less\; than\; or\; equal\; to\; this\; number.}$

分析：找x，大于等于num[]里面的k个元素,3个条件
1.如果k == 0且，num[1]>1（就是没有x大于num里面的0个元素）所以输入1，如果k==0且num[1] <= 1,说明
这时x还可以大于num里面的元素，与题意不符，所以输出-1。
2.k == n，x要大于等于num[]里面所有元素，题目中num[]最大为1000000000。
3.0 < k < n，只需要num[k] != num[k + 1]即可，因为这样就有k + 1个元素了，而题目只要k个元素

 

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                                                                        .                                                                             **/
                                                                        
#include <bits/stdc++.h>
using namespace std;

#define FF(i, a, b) for(int i = a; i < b; i++)
#define RR(i, a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define PR(x) printf("%d\n", x)
#define INF 0x3f3f3f3f
#define MAX 220000
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP; 

int main(void){
    #ifdef LOCAL
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
    #endif
    ios::sync_with_stdio(false); cin.tie(0);
    int i, n, k, num[MAX], ans, id;
    
    cin>>n>>k;
    for(i = 1; i <= n; i++) cin>>num[i];
    sort(num + 1, num + n + 1);
    if(k == 0){
        if(num[1] > 1) cout<<1<<endl;
        else cout<<-1<<endl;
    }
    else if(n == k) cout<<1000000000<<endl;
    else{
        if(num[k] == num[k + 1]) cout<<-1<<endl;
        else cout<<num[k]<<endl;
    }
    return EXIT_SUCCESS;
}

 

D. Divide by three, multiply by two

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

Copy

6
4 8 6 3 12 9

output

Copy

9 3 6 12 4 8 

input

Copy

4
42 28 84 126

output

Copy

126 42 84 28 

input

Copy

2
1000000000000000000 3000000000000000000

output

Copy

3000000000000000000 1000000000000000000 

Note

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

 

分析：题目好理解，后一个数是前一个数的 1/ 3，或者2倍,先统计所有数%3的个数，先按照这个数排列，再按照他们%2个数排列（或他们自身数大小排列，一样的，因为是2倍，后面数肯定大于前面数）

 

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                                                                        .                                                                             **/
                                                                        
#include <bits/stdc++.h>
using namespace std;

#define FF(i, a, b) for(int i = a; i < b; i++)
#define RR(i, a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define PR(x) printf("%d\n", x)
#define INF 0x3f3f3f3f
#define MAX 110
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP; 

struct Node{
    LL a;
    int b;
}num[MAX];



//ÔªËصÄÖØÅÅÁÐ  
bool cmp(const Node& a, const Node& b){
    if(a.b != b.b) return a.b > b.b;
    else return a.a < b.a;
}


int main(void){
    #ifdef LOCAL
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
    #endif
    ios::sync_with_stdio(false); cin.tie(0);
    int i, n; LL cnt;
    
    cin>>n;
    for(i = 0; i < n; i++){
        cin>>num[i].a;
        cnt = num[i].a;
        while(cnt % 3 == 0){cnt /= 3; num[i].b++;}
    }
    sort(num, num + n, cmp);
    for(i = 0; i < n; i++)    printf("%lld ", num[i].a);
    puts("");
    
    return EXIT_SUCCESS;
}

E. Cyclic Components

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

Copy

5 4
1 2
3 4
5 4
3 5

output

Copy

1

input

Copy

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.

分析：题意就是统计环的个数且这个环任意一点只能有2个边,简单DFS图论题目，2维数组统计每个点的度，条件如下：

1.该点度不为2，退出。

2.该点没访问过，标记，并访问与该点关联的所有点，ans++

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#include <bits/stdc++.h>
using namespace std;

#define FF(i, a, b) for(int i = a; i < b; i++)
#define RR(i, a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define PR(x) printf("%d\n", x)
#define INF 0x3f3f3f3f
#define MAX 220000
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP; 

vector<int>v[MAX];
int dis[MAX] = {0};
int flag;

void DFS(int u){
    dis[u] = 1;    //该点访问
    if(v[u].size() != 2) flag = false; //若改点的度不为2，则可以直接退出 
    for(int i = 0; i < v[u].size(); i++){ if(!dis[v[u][i]]) DFS(v[u][i]);} 
}



int main(void){
    #ifdef LOCAL
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
    #endif
    ios::sync_with_stdio(false); cin.tie(0);
    int i, j, n, m, a, b, ans;
    
    cin>>n>>m;ans = 0;
    for(i = 1; i <= m; i++){
        cin>>a>>b;
        //所有点的出入度都记录 
        v[a].push_back(b);
        v[b].push_back(a);
    }
    for(i = 1; i <= n; i++){
        if(!dis[i]){
            flag = 1;DFS(i);
            if(flag) ans++;
        }
    }
    cout<<ans<<endl;
    
    return EXIT_SUCCESS;
}

 

F. Consecutive Subsequence

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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:

• $[1,2,3,7,8,9]$

index = num[i],index++,输出i即可

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                                         :r;:::  ,                PH r,       ,;5: ,::::;7Ls7Lc7;:                                         ,:7JP17rJUs
                                                                  .:J: :;  .,:K6BQS7:.,.,.
                                                                    :r7Ji;r7;::;;.
                                                                        .                                                                             **/
                                                                        
#include <bits/stdc++.h>
using namespace std;

#define FF(i, a, b) for(int i = a; i < b; i++)
#define RR(i, a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define PR(x) printf("%d\n", x)
#define INF 0x3f3f3f3f
#define MAX 220000
#define MOD 1000000007
#define E 2.71828182845
#define M 8
#define N 6
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
vector<pair<string, int> > VP; 

map<LL, LL> mp;
int main(void){
    #ifdef LOCAL
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
    #endif
    ios::sync_with_stdio(false); cin.tie(0);
    int i, j, n;LL a[MAX], ans, index;
    
    cin>>n;ans = index = 0;
    for(i = 1; i <= n; i++){cin>>a[i]; mp[a[i]] = mp[a[i] - 1] + 1;}
    for(i = 1; i <= n; i++){
        if(mp[a[i]] > ans){
            ans = mp[a[i]];
            index = a[i];
        }
    }
    cout<<ans<<endl;
    index = index - ans + 1;
    for(i = 1; i <= n; i++){
        if(a[i] == index){
            printf("%d ", i);
            index++;
        }
    }
    puts("");
    
    return EXIT_SUCCESS;
}