Codeforces Round #481 (Div. 3)

A. Remove Duplicates

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Petya has an array $\mathrm{aa consisting\; of nn integers.\; He\; wants\; to\; remove\; duplicate\; (equal)\; elements.}$

Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed.

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 501\le n\le 50)\; —\; the\; number\; of\; elements\; in\; Petya\text{'}s\; array.}$

The following line contains a sequence ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 10001\le ai\le 1000)\; —\; the\; Petya\text{'}s\; array.}}^{}$

Output

In the first line print integer $\mathrm{xx —\; the\; number\; of\; elements\; which\; will\; be\; left\; in\; Petya\text{'}s\; array\; after\; he\; removed\; the\; duplicates.}$

In the second line print $\mathrm{xx integers\; separated\; with\; a\; space\; —\; Petya\text{'}s\; array\; after\; he\; removed\; the\; duplicates.\; For\; each\; unique\; element\; only\; the\; rightmost\; entry\; should\; be\; left.}$

Examples

input

Copy

6
1 5 5 1 6 1

output

Copy

3
5 6 1 

input

Copy

5
2 4 2 4 4

output

Copy

2
2 4 

input

Copy

5
6 6 6 6 6

output

Copy

1
6 

Note

In the first example you should remove two integers $\mathrm{11,\; which\; are\; in\; the\; positions 11 and 44.\; Also\; you\; should\; remove\; the\; integer 55,\; which\; is\; in\; the\; position 22.}$

In the second example you should remove integer $\mathrm{22,\; which\; is\; in\; the\; position 11,\; and\; two\; integers 44,\; which\; are\; in\; the\; positions 22 and 44.}$

In the third example you should remove four integers $\mathrm{66,\; which\; are\; in\; the\; positions 11, 22, 33 and 44.}$

$\mathrm{思路:水题；}$

代码：

#include<bits/stdc++.h>
using namespace std;
int flag[1005];
int a[55];
int b[55];
int top=0;
set<int> s;
int main(){
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
        s.insert(a[i]); 
    }
    printf("%d\n",s.size());
    for(int i=n-1;i>=0;i--){
        if(flag[a[i]]==0){
            flag[a[i]]=1;
            b[top++]=a[i];
        }
    }
    for(int i=top-1;i>=0;i--){
        if(i==top-1)
            printf("%d",b[i]);
        else    
            printf(" %d",b[i]);
    }
    printf("\n");
}

B. File Name

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You can not just take the file and send it. When Polycarp trying to send a file in the social network "Codehorses", he encountered an unexpected problem. If the name of the file contains three or more "x" (lowercase Latin letters "x") in a row, the system considers that the file content does not correspond to the social network topic. In this case, the file is not sent and an error message is displayed.

Determine the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. Print 0 if the file name does not initially contain a forbidden substring "xxx".

You can delete characters in arbitrary positions (not necessarily consecutive). If you delete a character, then the length of a string is reduced by $$$1$$$. For example, if you delete the character in the position $$$2$$$ from the string "exxxii", then the resulting string is "exxii".

Input

The first line contains integer $$$n$$$ $$$(3 \le n \le 100)$$$ — the length of the file name.

The second line contains a string of length $$$n$$$ consisting of lowercase Latin letters only — the file name.

Output

Print the minimum number of characters to remove from the file name so after that the name does not contain "xxx" as a substring. If initially the file name dost not contain a forbidden substring "xxx", print 0.

Examples

input

6
xxxiii

output

1

input

5
xxoxx

output

0

input

10
xxxxxxxxxx

output

8

Note

In the first example Polycarp tried to send a file with name contains number $$$33$$$, written in Roman numerals. But he can not just send the file, because it name contains three letters "x" in a row. To send the file he needs to remove any one of this letters.

思路：直接暴力；水题；

 代码：

#include<bits/stdc++.h>
using namespace std;
char s[105];
int main(){
    int n;
    scanf("%d",&n);
    scanf("%s",s);
    int len=strlen(s);
    int sum=0; 
    for(int i=0;i<=len-3;i++){
        int j;
        for(j=i;j<i+3;j++){
            if(s[j]!='x'){
                break;
            }
        }
        if(j==i+3) sum++;
    }
    printf("%d\n",sum);
}

C. Letters

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are $\mathrm{nn dormitories\; in\; Berland\; State\; University,\; they\; are\; numbered\; with\; integers\; from 11 to nn.\; Each\; dormitory\; consists\; of\; rooms,\; there\; are aiai rooms\; in ii-th\; dormitory.\; The\; rooms\; in ii-th\; dormitory\; are\; numbered\; from 11 to aiai.}$

A postman delivers letters. Sometimes there is no specific dormitory and room number in it on an envelope. Instead of it only a room number among all rooms of all $\mathrm{nn dormitories\; is\; written\; on\; an\; envelope.\; In\; this\; case,\; assume\; that\; all\; the\; rooms\; are\; numbered\; from 11 to a1+a2+\cdots +ana1+a2+\cdots +an and\; the\; rooms\; of\; the\; first\; dormitory\; go\; first,\; the\; rooms\; of\; the\; second\; dormitory\; go\; after\; them\; and\; so\; on.}$

For example, in case $\mathrm{n=2n=2, a1=3a1=3 and a2=5a2=5 an\; envelope\; can\; have\; any\; integer\; from 11 to 88 written\; on\; it.\; If\; the\; number 77 is\; written\; on\; an\; envelope,\; it\; means\; that\; the\; letter\; should\; be\; delivered\; to\; the\; room\; number 44 of\; the\; second\; dormitory.}$

For each of $\mathrm{mm letters\; by\; the\; room\; number\; among\; all nn dormitories,\; determine\; the\; particular\; dormitory\; and\; the\; room\; number\; in\; a\; dormitory\; where\; this\; letter\; should\; be\; delivered.}$

Input

The first line contains two integers $\mathrm{nn and mm (1\le n,m\le 2\cdot 105)(1\le n,m\le 2\cdot 105) —\; the\; number\; of\; dormitories\; and\; the\; number\; of\; letters.}$

The second line contains a sequence ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 1010)(1\le ai\le 1010),\; where aiai equals\; to\; the\; number\; of\; rooms\; in\; the ii-th\; dormitory.\; The\; third\; line\; contains\; a\; sequence b1,b2,\dots ,bmb1,b2,\dots ,bm (1\le bj\le a1+a2+\cdots +an)(1\le bj\le a1+a2+\cdots +an),\; where bjbj equals\; to\; the\; room\; number\; (among\; all\; rooms\; of\; all\; dormitories)\; for\; the jj-th\; letter.\; All bjbj are\; given\; in increasing order.}}^{}$

Output

Print $\mathrm{mm lines.\; For\; each\; letter\; print\; two\; integers ff and kk —\; the\; dormitory\; number ff (1\le f\le n)(1\le f\le n) and\; the\; room\; number kk in\; this\; dormitory (1\le k\le af)(1\le k\le af) to\; deliver\; the\; letter.}$

Examples

input

Copy

3 6
10 15 12
1 9 12 23 26 37

output

Copy

1 1
1 9
2 2
2 13
3 1
3 12

input

Copy

2 3
5 10000000000
5 6 9999999999

output

Copy

1 5
2 1
2 9999999994

Note

In the first example letters should be delivered in the following order:

• the first letter in room $\mathrm{11 of\; the\; first\; dormitory}$
• the second letter in room $\mathrm{99 of\; the\; first\; dormitory}$
• the third letter in room $\mathrm{22 of\; the\; second\; dormitory}$
• the fourth letter in room $\mathrm{1313 of\; the\; second\; dormitory}$
• the fifth letter in room $\mathrm{11 of\; the\; third\; dormitory}$
• the sixth letter in room $\mathrm{1212 of\; the\; third\; dormitory}$

$\mathrm{思路：先求前缀和；再暴力找；水题；}$

$\mathrm{代码：}$

 #include<bits/stdc++.h>
using namespace std;
long long a[200005];
long long b[200005];
int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++){
        scanf("%lld",&a[i]);
        if(i!=0)
            a[i]=a[i-1]+a[i];
    }
    for(int i=0;i<m;i++){
        scanf("%lld",&b[i]);
    }
    int t=0;
    for(int i=0;i<m;i++){
        if(b[i]<=a[t]){
            if(t==0){
                printf("%d %lld\n",t+1,b[i]);
            }
            else{
                printf("%d %lld\n",t+1,b[i]-a[t-1]);
            } 
        }
        else{
            while(b[i]>a[t]){
                t++;
            }
            if(t==0){
                printf("%d %lld\n",t+1,b[i]);
            }
            else{
                printf("%d %lld\n",t+1,b[i]-a[t-1]);
            } 
        }
    }
}