Educational Codeforces Round 42 (Rated for Div. 2)

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　A. Equator

Polycarp has created his own training plan to prepare for the programming contests. He will train for $\mathrm{nn\; days,\; all\; days\; are\; numbered\; from\; 11\; to\; nn,\; beginning\; from\; the\; first.}$

On the $\mathrm{ii-th\; day\; Polycarp\; will\; necessarily\; solve\; aiai\; problems.\; One\; evening\; Polycarp\; plans\; to\; celebrate\; the\; equator.\; He\; will\; celebrate\; it\; on\; the\; first\; evening\; of\; such\; a\; day\; that\; from\; the\; beginning\; of\; the\; training\; and\; to\; this\; day\; inclusive\; he\; will\; solve\; half\; or\; more\; of\; all\; the\; problems.}$

Determine the index of day when Polycarp will celebrate the equator.

Input

The first line contains a single integer $\mathrm{nn\; (1\le n\le 2000001\le n\le 200000)\; —\; the\; number\; of\; days\; to\; prepare\; for\; the\; programming\; contests.}$

The second line contains a sequence ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an\; (1\le ai\le 100001\le ai\le 10000),\; where\; aiai\; equals\; to\; the\; number\; of\; problems,\; which\; Polycarp\; will\; solve\; on\; the\; ii-th\; day.}}^{}$

Output

Print the index of the day when Polycarp will celebrate the equator.

注意奇数的一半和偶数的一半

#pragma warning(disable:4996)
#include<bitset>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;

const int maxn = 200005;

int n;
int a[maxn];

int main()
{
    while (cin >> n) {
        int tmp, sum = 0;
        for (int i = 1; i <= n; i++) {
            cin >> a[i];
            sum += a[i];
        }
        a[0] = 0;
        for (int i = 1; i <= n; i++) {
            a[i] += a[i - 1];
            if (sum % 2) {
                if (a[i] >= sum / 2 + 1) { cout << i << endl; break; }
            }
            else {
                if (a[i] >= sum / 2) { cout << i << endl; break; }
            }
        }
    }
    return 0;
}

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　B. Students in Railway Carriage

There are $\mathrm{nn\; consecutive\; seat\; places\; in\; a\; railway\; carriage.\; Each\; place\; is\; either\; empty\; or\; occupied\; by\; a\; passenger.}$

The university team for the Olympiad consists of $\mathrm{aa\; student-programmers\; and\; bb\; student-athletes.\; Determine\; the\; largest\; number\; of\; students\; from\; all\; a+ba+b\; students,\; which\; you\; can\; put\; in\; the\; railway\; carriage\; so\; that:}$

• no student-programmer is sitting next to the student-programmer;
• and no student-athlete is sitting next to the student-athlete.

In the other words, there should not be two consecutive (adjacent) places where two student-athletes or two student-programmers are sitting.

Consider that initially occupied seat places are occupied by jury members (who obviously are not students at all).

Input

The first line contain three integers $\mathrm{nn,\; aa\; and\; bb\; (1\le n\le 2\cdot 1051\le n\le 2\cdot 105,\; 0\le a,b\le 2\cdot 1050\le a,b\le 2\cdot 105,\; a+b>0a+b>0)\; —\; total\; number\; of\; seat\; places\; in\; the\; railway\; carriage,\; the\; number\; of\; student-programmers\; and\; the\; number\; of\; student-athletes.}$

The second line contains a string with length $\mathrm{nn,\; consisting\; of\; characters\; "."\; and\; "*".\; The\; dot\; means\; that\; the\; corresponding\; place\; is\; empty.\; The\; asterisk\; means\; that\; the\; corresponding\; place\; is\; occupied\; by\; the\; jury\; member.}$

Output

Print the largest number of students, which you can put in the railway carriage so that no student-programmer is sitting next to a student-programmer and no student-athlete is sitting next to a student-athlete.

题解：当前位置能放的话，如果前面没限制，则放最多的那类，否则，只能放另一类。

感受：我想到的是分类讨论，结果写不粗来，好菜啊！！！，在大佬的提示下，说这是很明显的贪心，随恍然大悟。结果又写挂了，忘了memset （use）!!!!!

#pragma warning(disable:4996)
#include<bitset>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;

const int INF = 2e9 + 7;
const int maxn = 200005;

int n, a, b;
int use[maxn];
char s[maxn];

int main()
{
    while (cin >> n >> a >> b) {
        scanf("%s", s);
        memset(use, 0, sizeof(use));
        for (int i = 0; i < n; i++) if (s[i] == '*') use[i] = -1;

        int t = 0;
        if (s[0] == '.') {
            if (a > b) {
                use[0] = 0;
                a--;
                t++;
            }
            else {
                use[0] = 1;
                b--;
                t++;
            }
        }

        for (int i = 1; i < n; i++) {
            if (use[i] == -1) continue;
            if (a == 0 && b == 0) break;
            if (use[i - 1] == -1) {
                if (a > b) {
                    use[i] = 0;
                    a--;
                    t++;
                }
                else {
                    use[i] = 1;
                    b--;
                    t++;
                }
            }
            else {
                if (use[i - 1] == 0) {
                    if (b) {
                        use[i] = 1;
                        b--;
                        t++;
                    }
                    else use[i] = -1;
                }
                else {
                    if (a) {
                        use[i] = 0;
                        a--;
                        t++;
                    }
                    else use[i] = -1;
                }
            }
        }

        cout << t << endl;
    }
    return 0;
}

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　C. Make a Square

You are given a positive integer $\mathrm{nn,\; written\; without\; leading\; zeroes\; (for\; example,\; the\; number\; 04\; is\; incorrect).}$

In one operation you can delete any digit of the given integer so that the result remains a positive integer without leading zeros.

Determine the minimum number of operations that you need to consistently apply to the given integer $\mathrm{nn\; to\; make\; from\; it\; the\; square\; of\; some\; positive\; integer\; or\; report\; that\; it\; is\; impossible.}$

An integer $\mathrm{xx\; is\; the\; square\; of\; some\; positive\; integer\; if\; and\; only\; if\; x=y2x=y2\; for\; some\; positive\; integer\; yy.}$

Input

The first line contains a single integer $\mathrm{nn\; (1\le n\le 2\cdot 1091\le n\le 2\cdot 109).\; The\; number\; is\; given\; without\; leading\; zeroes.}$

Output

If it is impossible to make the square of some positive integer from $\mathrm{nn,\; print\; -1.\; In\; the\; other\; case,\; print\; the\; minimal\; number\; of\; operations\; required\; to\; do\; it.}$

$\mathrm{题解：暴力寻找y，然后直接匹配就行了。}$

#pragma warning(disable:4996)
#include<bitset>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;

const int INF = 2e9 + 7;
const int maxn = 100005;

int n, m;
int a[105], b[105];

int main()
{
    while (cin >> n) {
        int p = 0, q = 0, t = 0;
        int k = n;
        while (k) {
            b[q++] = k % 10;
            k = k / 10;
        }
    
        int ans = INF;
        for (int i = 1; i*i <= n; i++) {

            p = 0;
            int tmp = i * i;
            while (tmp) {
                a[p++] = tmp % 10;
                tmp = tmp / 10;
            }

            t = 0;
            for (int j = 0; j < q; j++) if (b[j] == a[t]) {
                t++;
                if (t == p) break;
            }
            if (t == p) ans = min(ans, q - p);
        }
        if (ans == INF) cout << "-1" << endl;
        else cout << ans << endl;
    }
    return 0;
}

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　D. Merge Equals

You are given an array of positive integers. While there are at least two equal elements, we will perform the following operation. We choose the smallest value $\mathrm{xx\; that\; occurs\; in\; the\; array\; 22\; or\; more\; times.\; Take\; the\; first\; two\; occurrences\; of\; xx\; in\; this\; array\; (the\; two\; leftmost\; occurrences).\; Remove\; the\; left\; of\; these\; two\; occurrences,\; and\; the\; right\; one\; is\; replaced\; by\; the\; sum\; of\; this\; two\; values\; (that\; is,\; 2\cdot x2\cdot x\; ).}$

Determine how the array will look after described operations are performed.

For example, consider the given array looks like $[3,4,1,2,2,1,1][3,4,1,2,2,1,1]\; .\; It\; will\; be\; changed\; in\; the\; following\; way:\; [3,4,1,2,2,1,1] \to [3,4,2,2,2,1] \to [3,4,4,2,1] \to [3,8,2,1][3,4,1,2,2,1,1] \to [3,4,2,2,2,1] \to [3,4,4,2,1] \to [3,8,2,1]\; .$

If the given array is look like $[1,1,3,1,1][1,1,3,1,1]\; it\; will\; be\; changed\; in\; the\; following\; way:\; [1,1,3,1,1] \to [2,3,1,1] \to [2,3,2] \to [3,4][1,1,3,1,1] \to [2,3,1,1] \to [2,3,2] \to [3,4]\; .$

Input

The first line contains a single integer $\mathrm{nn\; (2\le n\le 1500002\le n\le 150000\; )\; —\; the\; number\; of\; elements\; in\; the\; array.}$

The second line contains a sequence from $\mathrm{nn\; elements\; a1,a2,\dots ,ana1,a2,\dots ,an\; (1\le ai\le 1091\le ai\le 109\; )\; —\; the\; elements\; of\; the\; array.}$

Output

In the first line print an integer $\mathrm{kk\; —\; the\; number\; of\; elements\; in\; the\; array\; after\; all\; the\; performed\; operations.\; In\; the\; second\; line\; print\; kk\; integers\; —\; the\; elements\; of\; the\; array\; after\; all\; the\; performed\; operations.}$

Examples

Input

Copy

7
3 4 1 2 2 1 1

Output

Copy

4
3 8 2 1 

Input

Copy

5
1 1 3 1 1

Output

Copy

2
3 4 

Input

Copy

5
10 40 20 50 30

Output

Copy

5
10 40 20 50 30 

Note

The first two examples were considered in the statement.

In the third example all integers in the given array are distinct, so it will not change.

题解：用map记录每个数的位置，如果当前数已经出现过，则消掉当前数的位置，同时数乘以2。还要记录总共消掉了多少个数。

感受：过了这么久才补，一直想不出怎么做。。。只有看题解了，兰儿题解也看不懂。。。然后去看别人的代码。。。这写法好清新脱俗啊，易懂又好写，就它了。不过python的代码挺有意思的，可以学习一波。

#pragma warning(disable:4996)
#include<map>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long 
using namespace std;

const int maxn = 200005;

int n;
ll a[maxn];

int main()
{
    while (scanf("%d", &n) != EOF) {
        map<ll,int> mp;

        int cnt = 0;
        for (int i = 1; i <= n; i++) {
            scanf("%I64d", a + i);
            while (mp[a[i]]) {
                a[mp[a[i]]] = 0;     //将第一次出现的数置0，为了输出方便
                mp[a[i]] = 0;        //重点，将这个数出现的位置消除掉（即置0），是避免以后出现相同的数，重复计算，比如4 2 2 2
                a[i] = a[i] * 2;
                cnt++;
            }
            mp[a[i]] = i;
        }

        printf("%d
", n - cnt);
        for (int i = 1; i <= n; i++) if (a[i]) printf("%I64d ", a[i]);
        printf("
");
    }
    return 0;
}