Codeforces Round #479 (Div. 3)

手速场2333，这群人贼牛逼！手速贼快！

 

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

---

水题，不说。

 1 #include<bits/stdc++.h>
 2 #define clr(x) memset(x,0,sizeof(x))
 3 #define clr_1(x) memset(x,-1,sizeof(x))
 4 #define clrmax(x) memset(x,0x3f3f3f3f,sizeof(x))
 5 #define mod 1000000007
 6 #define INF 0x3f3f3f3f
 7 #define LL long long
 8 #define pb push_back
 9 #define pbk pop_back
10 #define ls(i) (i<<1)
11 #define rs(i) (i<<1|1)
12 #define mp make_pair
13 using namespace std;
14 LL n;
15 int k;
16 int main()
17 {
18     scanf("%I64d%d",&n,&k);
19     for(int i=1;i<=k;i++)
20     {
21         if(n%10==0)
22             n/=10;
23         else
24             n--;
25     }
26     printf("%I64d\n",n);
27     return 0;
28 }

View Code

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

---

拿个string 和map 存一存就好了。

 1 #include<bits/stdc++.h>
 2 #define clr(x) memset(x,0,sizeof(x))
 3 #define clr_1(x) memset(x,-1,sizeof(x))
 4 #define clrmax(x) memset(x,0x3f3f3f3f,sizeof(x))
 5 #define mod 1000000007
 6 #define INF 0x3f3f3f3f
 7 #define LL long long
 8 #define pb push_back
 9 #define pbk pop_back
10 #define ls(i) (i<<1)
11 #define rs(i) (i<<1|1)
12 #define mp make_pair
13 using namespace std;
14 const int N=1e2+10;
15 map<string,int> st;
16 string str,s;
17 int n,ans;
18 int main()
19 {
20     cin>>n;
21     cin>>s;
22     ios::sync_with_stdio(false);
23     for(int i=0;i<n-1;i++)
24     {
25         str="";
26         str+=s[i];
27         str+=s[i+1];
28         st[str]++;
29     }
30     ans=0;
31     for(auto p:st)
32     {
33         if(p.second>ans)
34         {
35             ans=p.second;
36             str=p.first;
37         }
38     }
39     cout<<str<<endl;
40 }

View Code

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

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

---

如果有解就是排完序后第k个。特判k==0，当最小数为1的时候无解，否则为1。特判k>n无解。特判a[k]==a[k+1]的情况，这样也是无解的。

 1 #include<bits/stdc++.h>
 2 #define clr(x) memset(x,0,sizeof(x))
 3 #define clr_1(x) memset(x,-1,sizeof(x))
 4 #define clrmax(x) memset(x,0x3f3f3f3f,sizeof(x))
 5 #define mod 1000000007
 6 #define INF 0x3f3f3f3f
 7 #define LL long long
 8 #define pb push_back
 9 #define pbk pop_back
10 #define ls(i) (i<<1)
11 #define rs(i) (i<<1|1)
12 #define mp make_pair
13 using namespace std;
14 const int N=2e5+10;
15 int a[N];
16 int n,m,k,t;
17 int main()
18 {
19     scanf("%d%d",&n,&k);
20     for(int i=1;i<=n;i++)
21         scanf("%d",a+i);
22     sort(a+1,a+n+1);
23     if(k==0)
24     {
25         if(a[1]==1)
26             printf("-1\n");
27         else
28             printf("1\n");
29         return 0;
30     }
31     printf("%d\n",(k>n || (k<n && a[k]==a[k+1]))?-1:a[k]);
32     return 0;
33 }

View Code

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 

---

我可能是代码最长的那个233333。
反正就暴力链嘛，找到一个链一个，然后找没有被链的那个当头输出就好了。

 1 #include<bits/stdc++.h>
 2 #define clr(x) memset(x,0,sizeof(x))
 3 #define clr_1(x) memset(x,-1,sizeof(x))
 4 #define clrmax(x) memset(x,0x3f3f3f3f,sizeof(x))
 5 #define mod 1000000007
 6 #define INF 0x3f3f3f3f
 7 #define LL long long
 8 #define pb push_back
 9 #define pbk pop_back
10 #define ls(i) (i<<1)
11 #define rs(i) (i<<1|1)
12 #define mp make_pair
13 using namespace std;
14 const int N=1e2+10;
15 LL a[N];
16 int nexted[N];
17 int n,hd,vis[N];
18 int main()
19 {
20     scanf("%d",&n);
21     for(int i=1;i<=n;i++)
22         scanf("%I64d",a+i);
23     for(int i=1;i<=n;i++)
24         for(int j=1;j<=n;j++)
25             if(!vis[j] && a[j]%2==0 && a[i]==a[j]/2)
26             {
27                 nexted[i]=j;
28                 vis[j]=1;
29                 break;
30             }
31             else if(!vis[j] && a[i]%3==0 && a[i]==a[j]*3)
32             {
33                 nexted[i]=j;
34                 vis[j]=1;
35                 break;
36             }
37     for(int i=1;i<=n;i++)
38         if(vis[i]==0)
39         {
40             hd=i;
41             break;
42         }
43     for(int i=1;i<=n;i++)
44     {
45         printf("%I64d ",a[hd]);
46         hd=nexted[hd];
47     }
48     printf("\n");
49     return 0;
50 }

View Code

然后听说了排序新奇写法，就是把含3因数多的尽可能的往前排，在含3因数一样的情况下小的数排前面。这样因为保证有解，排序出来的结果就是答案。

学了下lambda表达式

#include<bits/stdc++.h>
using namespace std;
long long n,a[110];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%I64d",a+i);
    sort(a+1,a+n+1,[](long long a,long long b)
         {
             int cnt[2]={0,0};
             while(a%3==0) a/=3,cnt[0]++;
             while(b%3==0) b/=3,cnt[1]++;
             return cnt[0]==cnt[1]?a<b:cnt[0]>cnt[1];
         });
    for(int i=1;i<=n;i++)
        printf("%I64d%c",a[i]," \n"[i==n]);
    return 0;
}

View Code

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

---

对每个点都做个标记。每个点都dfs一下判断是不是形成一个圈，如果度>2的就可以直接return了。

 1 #include<bits/stdc++.h>
 2 #define clr(x) memset(x,0,sizeof(x))
 3 #define clr_1(x) memset(x,-1,sizeof(x))
 4 #define clrmax(x) memset(x,0x3f3f3f3f,sizeof(x))
 5 #define mod 1000000007
 6 #define INF 0x3f3f3f3f
 7 #define LL long long
 8 #define pb push_back
 9 #define pbk pop_back
10 #define ls(i) (i<<1)
11 #define rs(i) (i<<1|1)
12 #define mp make_pair
13 using namespace std;
14 const int N=2e5+10;
15 struct edg
16 {
17     int next,to;
18 }edge[N<<2];
19 int head[N],etot,vis[N],ct[N];
20 int n,m,u,v;
21 int ans;
22 void init()
23 {
24     clr_1(head);
25     clr(vis);
26     clr(ct);
27     etot=0;
28     ans=0;
29     return ;
30 }
31 void addedge(int u,int v)
32 {
33     edge[++etot]=(edg){head[u],v};
34     head[u]=etot;
35     return ;
36 }
37 bool dfs(int u,int hed,int fa,int dep)
38 {
39     vis[u]=1;
40     if(ct[u]>2)
41         return 0;
42     int p;
43     for(int i=head[u];i!=-1;i=edge[i].next)
44     {
45         p=edge[i].to;
46         if(p==fa)
47             continue;
48         if(vis[p] && p==hed)
49         {
50             if(dep>=2)
51                 return 1;
52             else
53                 return 0;
54         }
55         if(!vis[p])
56             return dfs(p,hed,u,dep+1);
57     }
58     return 0;
59 }
60 int main()
61 {
62     scanf("%d%d",&n,&m);
63     init();
64     for(int i=1;i<=m;i++)
65     {
66         scanf("%d%d",&u,&v);
67         addedge(u,v);
68         addedge(v,u);
69         ct[u]++;
70         ct[v]++;
71     }
72     for(int i=1;i<=n;i++)
73     {
74         if(!vis[i])
75             if(dfs(i,i,i,0))
76                 ans++;
77     }
78     printf("%d\n",ans);
79     return 0;
80 }

View Code

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 

---

LIS的 O(n) 写法 加上离散化。拿上lower_bound去找离散化的数。这样就能解决了。不过map的离散化代码能更短点。

 1 #include<bits/stdc++.h>
 2 #define clr(x) memset(x,0,sizeof(x))
 3 #define clr_1(x) memset(x,-1,sizeof(x))
 4 #define clrmax(x) memset(x,0x3f3f3f3f,sizeof(x))
 5 #define mod 1000000007
 6 #define INF 0x3f3f3f3f
 7 #define LL long long
 8 #define pb push_back
 9 #define pbk pop_back
10 #define ls(i) (i<<1)
11 #define rs(i) (i<<1|1)
12 using namespace std;
13 const int N=2e5+10;
14 int a[N],uni[N],n,m,now[N],pre[N],last,maxn,t,p,ans[N],num[N];
15 int main()
16 {
17     scanf("%d",&n);
18     for(int i=1;i<=n;i++)
19     {
20         scanf("%d",a+i);
21         uni[i]=a[i];
22     }
23     sort(uni+1,uni+n+1);
24     m=unique(uni+1,uni+n+1)-uni-1;
25     last=maxn=1;
26     for(int i=1;i<=n;i++)
27     {
28         t=lower_bound(uni+1,uni+m+1,a[i])-uni;
29         if(num[t]<1)
30         {
31             now[t]=i;
32             num[t]=1;
33         }
34         p=lower_bound(uni+1,uni+m+1,a[i]-1)-uni;
35         if(p>m || uni[p]!=a[i]-1)
36             continue;
37         if(num[t]<num[p]+1)
38         {
39             now[t]=i;
40             num[t]=num[p]+1;
41             pre[i]=now[p];
42         }
43         if(num[t]>maxn)
44         {
45             last=i;
46             maxn=num[t];
47         }
48     }
49     printf("%d\n",maxn);
50     for(int i=maxn;i>=1;i--)
51     {
52         ans[i]=last;
53         last=pre[last];
54     }
55     for(int i=1;i<=maxn;i++)
56         printf("%d ",ans[i]);
57     printf("\n");
58     return 0;
59 }

View Code