Codeforces Round #496 (Div. 3) ABCDE

A. Tanya and Stairways

Little girl Tanya climbs the stairs inside a multi-storey building. Every time Tanya climbs a stairway, she starts counting steps from $\mathrm{11 to\; the\; number\; of\; steps\; in\; this\; stairway.\; She\; speaks\; every\; number\; aloud.\; For\; example,\; if\; she\; climbs\; two\; stairways,\; the\; first\; of\; which\; contains 33steps,\; and\; the\; second\; contains 44 steps,\; she\; will\; pronounce\; the\; numbers 1,2,3,1,2,3,41,2,3,1,2,3,4.}$

You are given all the numbers pronounced by Tanya. How many stairways did she climb? Also, output the number of steps in each stairway.

The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.

Input

The first line contains $\mathrm{nn (1\le n\le 10001\le n\le 1000)\; —\; the\; total\; number\; of\; numbers\; pronounced\; by\; Tanya.}$

The second line contains integers ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 10001\le ai\le 1000)\; —\; all\; the\; numbers\; Tanya\; pronounced\; while\; climbing\; the\; stairs,\; in\; order\; from\; the\; first\; to\; the\; last\; pronounced\; number.\; Passing\; a\; stairway\; with xx steps,\; she\; will\; pronounce\; the\; numbers 1,2,\dots ,x1,2,\dots ,x in\; that\; order.}}^{}$

The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.

Output

In the first line, output $\mathrm{tt —\; the\; number\; of\; stairways\; that\; Tanya\; climbed.\; In\; the\; second\; line,\; output tt numbers\; —\; the\; number\; of\; steps\; in\; each\; stairway\; she\; climbed.\; Write\; the\; numbers\; in\; the\; correct\; order\; of\; passage\; of\; the\; stairways.}$

Examples

input

Copy

7
1 2 3 1 2 3 4

output

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2
3 4 

input

Copy

4
1 1 1 1

output

Copy

4
1 1 1 1 

input

Copy

5
1 2 3 4 5

output

Copy

1
5 

input

Copy

5
1 2 1 2 1

output

Copy

3
2 2 1 

签到题

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1010;
int a[N];

int main() {
    int n;
    cin >> n;
    std::vector<int> v;
    for(int i = 1; i <= n; i ++) cin >> a[i];
    int tmp = 1;
    for(int i = 2; i <= n; i ++) {
        if(a[i] == 1) v.push_back(a[i-1]);
    }
    v.push_back(a[n]);
    cout << v.size() << endl;
    for(int i = 0; i < v.size(); i ++) {
        cout << v[i] << ' ';
    }

    return 0;
}

 

B. Delete from the Left

You are given two strings $\mathrm{ss and tt.\; In\; a\; single\; move,\; you\; can\; choose\; any\; of\; two\; strings\; and\; delete\; the\; first\; (that\; is,\; the\; leftmost)\; character.\; After\; a\; move,\; the\; length\; of\; the\; string\; decreases\; by 11.\; You\; can\text{'}t\; choose\; a\; string\; if\; it\; is\; empty.}$

For example:

• by applying a move to the string "where", the result is the string "here",
• by applying a move to the string "a", the result is an empty string "".

You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.

Write a program that finds the minimum number of moves to make two given strings $\mathrm{ss and tt equal.}$

Input

The first line of the input contains $\mathrm{ss.\; In\; the\; second\; line\; of\; the\; input\; contains tt.\; Both\; strings\; consist\; only\; of\; lowercase\; Latin\; letters.\; The\; number\; of\; letters\; in\; each\; string\; is\; between\; 1\; and 2\cdot 1052\cdot 105,\; inclusive.}$

Output

Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.

Examples

input

Copy

test
west

output

Copy

2

input

Copy

codeforces
yes

output

Copy

9

input

Copy

test
yes

output

Copy

7

input

Copy

b
ab

output

Copy

1

Note

In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".

In the second example, the move should be applied to the string "codeforces" $\mathrm{88 times.\; As\; a\; result,\; the\; string\; becomes\; "codeforces" \to \to "es".\; The\; move\; should\; be\; applied\; to\; the\; string\; "yes"\; once.\; The\; result\; is\; the\; same\; string\; "yes" \to \to "es".}$

In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.

In the fourth example, the first character of the second string should be deleted.

求从右边开始相同的数量是多少。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 2e5+10;
char s[N], t[N];

int main() {
    cin >> s >> t;
    int lens = strlen(s), lent = strlen(t);
    int i = lens-1, j = lent-1;
    int ans = 0;
    while(s[i] == t[j] && i >= 0 && j >= 0) {
        i--;j--;
        ans += 2;
    }
    cout << lens+lent-ans << endl;
    return 0;
}

C. Summarize to the Power of Two

A sequence ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an is\; called\; good\; if,\; for\; each\; element aiai,\; there\; exists\; an\; element ajaj (i\ne ji\ne j)\; such\; that ai+ajai+aj is\; a\; power\; of\; two\; (that\; is, 2d2d for\; some\; non-negative\; integer dd).}}^{}$

For example, the following sequences are good:

• $[5,3,11][5,3,11] (for\; example,\; for a1=5a1=5 we\; can\; choose a2=3a2=3.\; Note\; that\; their\; sum\; is\; a\; power\; of\; two.\; Similarly,\; such\; an\; element\; can\; be\; found\; for a2a2 and a3a3),$
• $[1,1,1,1023][1,1,1,1023],$
• $[7,39,89,25,89][7,39,89,25,89],$
• $[][].$

Note that, by definition, an empty sequence (with a length of $\mathrm{00)\; is\; good.}$

For example, the following sequences are not good:

• $[16][16] (for a1=16a1=16,\; it\; is\; impossible\; to\; find\; another\; element ajaj such\; that\; their\; sum\; is\; a\; power\; of\; two),$
• $[4,16][4,16] (for a1=4a1=4,\; it\; is\; impossible\; to\; find\; another\; element ajaj such\; that\; their\; sum\; is\; a\; power\; of\; two),$
• $[1,3,2,8,8,8][1,3,2,8,8,8] (for a3=2a3=2,\; it\; is\; impossible\; to\; find\; another\; element ajaj such\; that\; their\; sum\; is\; a\; power\; of\; two).$

You are given a sequence ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an.\; What\; is\; the\; minimum\; number\; of\; elements\; you\; need\; to\; remove\; to\; make\; it\; good?\; You\; can\; delete\; an\; arbitrary\; set\; of\; elements.}}^{}$

Input

The first line contains the integer $\mathrm{nn (1\le n\le 1200001\le n\le 120000)\; —\; the\; length\; of\; the\; given\; sequence.}$

The second line contains the sequence of integers ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 1091\le ai\le 109).}}^{}$

Output

Print the minimum number of elements needed to be removed from the given sequence in order to make it good. It is possible that you need to delete all $\mathrm{nn elements,\; make\; it\; empty,\; and\; thus\; get\; a\; good\; sequence.}$

Examples

input

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6
4 7 1 5 4 9

output

Copy

1

input

Copy

5
1 2 3 4 5

output

Copy

2

input

Copy

1
16

output

Copy

1

input

Copy

4
1 1 1 1023

output

Copy

0

Note

In the first example, it is enough to delete one element ${\mathrm{a4=5a4=5.\; The\; remaining\; elements\; form\; the\; sequence [4,7,1,4,9][4,7,1,4,9],\; which\; is\; good}}^{}$

找两个能够成2^d的数。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 2e5+10;
int a[N], b[33];
map<int,int> mp;
int main() {
    int n, x;
    for(int i = 0; i < 31; i ++) b[i] = (1<<i);
    cin >> n;
    for(int i = 1; i <= n; i ++) {
        scanf("%d", &a[i]);
        mp[a[i]]++;
    }
    int ans = 0;
    map<int,int>::iterator it = mp.begin();
    for(; it != mp.end(); it ++) {
        bool flag = false;
        for(int j = 30; j > 0; j --) {
            int tmp = b[j] - (*it).first;
            if(tmp <= 0) break;
            if(mp.count(tmp)) {
                if(tmp == (*it).first && mp[tmp] == 1) continue;
                // cout << (*it).first << ' '<< tmp << endl;
                flag = true;
                break;
            }
        }
        if(!flag) {
            ans += (*it).second;
        }
    }
    cout << ans << endl;
    return 0;
}

 

D. Polycarp and Div 3

Polycarp likes numbers that are divisible by 3.

He has a huge number $\mathrm{ss.\; Polycarp\; wants\; to\; cut\; from\; it\; the\; maximum\; number\; of\; numbers\; that\; are\; divisible\; by 33.\; To\; do\; this,\; he\; makes\; an\; arbitrary\; number\; of\; vertical\; cuts\; between\; pairs\; of\; adjacent\; digits.\; As\; a\; result,\; after mm such\; cuts,\; there\; will\; be m+1m+1 parts\; in\; total.\; Polycarp\; analyzes\; each\; of\; the\; obtained\; numbers\; and\; finds\; the\; number\; of\; those\; that\; are\; divisible\; by 33.}$

For example, if the original number is $\mathrm{s=3121s=3121,\; then\; Polycarp\; can\; cut\; it\; into\; three\; parts\; with\; two\; cuts: 3|1|213|1|21.\; As\; a\; result,\; he\; will\; get\; two\; numbers\; that\; are\; divisible\; by 33.}$

Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.

What is the maximum number of numbers divisible by $\mathrm{33 that\; Polycarp\; can\; obtain?}$

Input

The first line of the input contains a positive integer $\mathrm{ss.\; The\; number\; of\; digits\; of\; the\; number ss is\; between 11 and 2\cdot 1052\cdot 105,\; inclusive.\; The\; first\; (leftmost)\; digit\; is\; not\; equal\; to 0.}$

Output

Print the maximum number of numbers divisible by $\mathrm{33 that\; Polycarp\; can\; get\; by\; making\; vertical\; cuts\; in\; the\; given\; number ss.}$

Examples

input

Copy

3121

output

Copy

2

input

Copy

6

output

Copy

1

input

Copy

1000000000000000000000000000000000

output

Copy

33

input

Copy

201920181

output

Copy

4

Note

In the first example, an example set of optimal cuts on the number is 3|1|21.

In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $33.$

In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $\mathrm{3333 digits 0.\; Each\; of\; the 3333digits 0 forms\; a\; number\; that\; is\; divisible\; by 33.}$

In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $\mathrm{00, 99, 201201 and 8181 are\; divisible\; by 33.}$

最多能分成多少份总和模3等于0 的数

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 2e5+10;
char str[N];
int main() {
    scanf("%s", str);
    int len = strlen(str);
    int ans = 0, tmp = 0, sum = 0, n = 0;
    for(int i = 0; i < len; i ++) {
        tmp = (str[i]-'0') % 3;
        sum += tmp;
        n++;
        if(tmp == 0 || sum%3==0 || n == 3) {
            ans++;
            sum = n = 0;
        }
    }
    cout << ans << endl;
    return 0;
}

 

D. Polycarp and Div 3

Polycarp likes numbers that are divisible by 3.

He has a huge number $\mathrm{ss.\; Polycarp\; wants\; to\; cut\; from\; it\; the\; maximum\; number\; of\; numbers\; that\; are\; divisible\; by 33.\; To\; do\; this,\; he\; makes\; an\; arbitrary\; number\; of\; vertical\; cuts\; between\; pairs\; of\; adjacent\; digits.\; As\; a\; result,\; after mm such\; cuts,\; there\; will\; be m+1m+1 parts\; in\; total.\; Polycarp\; analyzes\; each\; of\; the\; obtained\; numbers\; and\; finds\; the\; number\; of\; those\; that\; are\; divisible\; by 33.}$

For example, if the original number is $\mathrm{s=3121s=3121,\; then\; Polycarp\; can\; cut\; it\; into\; three\; parts\; with\; two\; cuts: 3|1|213|1|21.\; As\; a\; result,\; he\; will\; get\; two\; numbers\; that\; are\; divisible\; by 33.}$

Polycarp can make an arbitrary number of vertical cuts, where each cut is made between a pair of adjacent digits. The resulting numbers cannot contain extra leading zeroes (that is, the number can begin with 0 if and only if this number is exactly one character '0'). For example, 007, 01 and 00099 are not valid numbers, but 90, 0 and 10001 are valid.

What is the maximum number of numbers divisible by $\mathrm{33 that\; Polycarp\; can\; obtain?}$

Input

The first line of the input contains a positive integer $\mathrm{ss.\; The\; number\; of\; digits\; of\; the\; number ss is\; between 11 and 2\cdot 1052\cdot 105,\; inclusive.\; The\; first\; (leftmost)\; digit\; is\; not\; equal\; to 0.}$

Output

Print the maximum number of numbers divisible by $\mathrm{33 that\; Polycarp\; can\; get\; by\; making\; vertical\; cuts\; in\; the\; given\; number ss.}$

Examples

input

Copy

3121

output

Copy

2

input

Copy

6

output

Copy

1

input

Copy

1000000000000000000000000000000000

output

Copy

33

input

Copy

201920181

output

Copy

4

Note

In the first example, an example set of optimal cuts on the number is 3|1|21.

In the second example, you do not need to make any cuts. The specified number 6 forms one number that is divisible by $33.$

In the third example, cuts must be made between each pair of digits. As a result, Polycarp gets one digit 1 and $\mathrm{3333 digits 0.\; Each\; of\; the 3333digits 0 forms\; a\; number\; that\; is\; divisible\; by 33.}$

In the fourth example, an example set of optimal cuts is 2|0|1|9|201|81. The numbers $\mathrm{00, 99, 201201 and 8181 are\; divisible\; by 33.}$

 

找到m的位置，然后计算左边大于m的数与小于m的数的差值。接着从右边反向计算算出结果。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 2e5+10;
int a[N], n, m, k;
map<int,ll> mp;
int main() {
    scanf("%d %d", &n, &m);
    for(int i = 1; i <= n; i ++) {
        scanf("%d", &a[i]);
        if(a[i] == m) k = i;
    }
    int s = 1;
    for(int i = k; i > 0; i --) {
        if(a[i] > m) s++;
        else s--;
        mp[s]++;
    }
    s = -1;
    ll ans = 0;
    for(int i = k; i <= n; i ++) {
        if(a[i] > m) s--;
        else s++;
        ans += mp[s] + mp[s+1];
    }
    printf("%lld\n",ans);
    return 0;
}