Codeforces Round #490 (Div. 3)

A、Mishka and Contest

思路：简单贪心。每次删掉首尾不超过k的元素，直到分别第一次遇到超过k的元素就不再继续删除即可。

AC代码：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <complex>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <list>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // 上右下左
const int mx[8] = {-1, -2, -2, -1, 1, 2, 2, 1}; // 马可走的八个方向
const int my[8] = {-2, -1, 1, 2, 2, 1, -1, -2};
const double eps = 1e-6;
const double PI = acos(-1.0);
const int maxn = 105;
const int inf = 0x3f3f3f3f;

int n, k, ans, a[maxn];

int main() {
    while(cin >> n >> k) {
        ans = 0;
        for(int i = 0; i < n; ++i) cin >> a[i];
        for(int i = 0, j = n - 1; i <= j;) {
            if(a[i] <= k) ++i, ++ans;
            else if(a[j] <= k) --j, ++ans;
            else break;
        }
        cout << ans << endl;
    }
    return 0;
}

B、Reversing Encryption

思路：暴力。简单按照n的所有公约数从小到大有序反转一下即可。

AC代码：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <complex>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <list>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // 上右下左
const int mx[8] = {-1, -2, -2, -1, 1, 2, 2, 1}; // 马可走的八个方向
const int my[8] = {-2, -1, 1, 2, 2, 1, -1, -2};
const double eps = 1e-6;
const double PI = acos(-1.0);
const int maxn = 1e5+5;
const int inf = 0x3f3f3f3f;

int n;
string str;

int main() {
    while(cin >> n >> str) {
        for(int i = 1; i <= n; ++i)
            if(n % i == 0) reverse(str.begin(), str.begin() + i);
        cout << str << endl;
    }
    return 0;
}

C、Alphabetic Removals

思路：暴力。关联式容器multimap（键---值：一对多）的简单应用。注意：map.erase(迭代器)的结果只是被删元素的迭代器失效，但因为其返回值为void，所以要采用map.erase(it++)的方式删除被删元素的迭代器，此时it已经指向下一个迭代器的位置了。

AC代码：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <complex>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <list>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // 上右下左
const int mx[8] = {-1, -2, -2, -1, 1, 2, 2, 1}; // 马可走的八个方向
const int my[8] = {-2, -1, 1, 2, 2, 1, -1, -2};
const double eps = 1e-6;
const double PI = acos(-1.0);
const int maxn = 4e5+5;
const int inf = 0x3f3f3f3f;

int n, k, len;

multimap<char, int> mp1;
multimap<char, int>::iterator it1;

map<int, char> mp2;
map<int, char>::iterator it2;

char str[maxn];

int main() {
    while(cin >> n >> k >> str) {
        len = strlen(str);
        if(k == len) continue;
        mp1.clear(); mp2.clear();
        for(int i = 0; i < len; ++i) mp1.insert(make_pair(str[i], i));
        // 注意：mp1.erase(迭代器)的结果只是被删元素的迭代器失效，但因为其返回值为void，所以要采用erase(it++)的方式删除被删元素的迭代器
        for(it1 = mp1.begin(); k && (it1 != mp1.end()); --k) mp1.erase(it1++);
        for(it1 = mp1.begin(); it1 != mp1.end(); ++it1) mp2[it1 -> second] = it1 -> first;
        for(it2 = mp2.begin(); it2 != mp2.end(); ++it2) printf("%c", it2 -> second);
        puts("");
    }
    return 0;
}