Codeforces Round #726 (Div. 2)

A. Arithmetic Array

题意：

题解：

代码：

#include <bits/stdc++.h>

#define int long long
#define debug(x) cout << #x << " = " << x << endl;
using namespace std;

inline int rd() {
    
   int s = 0, w = 1;
   char ch = getchar();
   while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1; ch = getchar();}
   while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
   return s * w;
}

int const MAXN = 2e5 + 10;
int n, m, T, a[MAXN];

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

B. Bad Boy

题意：

题解：

代码：

#include <bits/stdc++.h>

#define int long long
#define debug(x) cout << #x << " = " << x << endl;
using namespace std;

inline int rd() {
    
   int s = 0, w = 1;
   char ch = getchar();
   while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1; ch = getchar();}
   while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
   return s * w;
}

int const MAXN = 2e5 + 10;
int n, m, T;
int x[10], y[10];

signed main() {
    
    cin >> T;
    while(T--) {
    
        int x1, y1;
        cin >> n >> m >> x1 >> y1;
        x[0] = 1, y[0] = 1;
        x[1] = 1, y[1] = m;
        x[2] = n, y[2] = 1;
        x[3] = n, y[3] = m;
        int Max = -1;
        int resx1 = 0, resy1 = 0;
        int resx2 = 0, resy2 = 0;
        for (int i = 0; i <=3; ++i) {
    
            for (int j = 0; j <= 3; ++j) {
    
                // if (i == j) continue;
                // if (x[i] == x1 && y[i] == y1) continue;
                // if (x[j] == x1 && y[j] == y1) continue;
                int tmp = abs(x[i] - x1) + abs(y[i] - y1) + abs(x[i] - x[j]) + abs(y[i] - y[j]) + abs(x[j] - x1) + abs(y[j] - y1);
                // cout << x[i] << " " << y[i] << " " << x[j] << " " << y[j] << " " << tmp << endl;
                if (tmp > Max) {
    
                    Max = tmp;
                    
                    resx1 = x[i], resy1 = y[i];
                    resx2 = x[j], resy2 = y[j];
                }
            }
        }
        cout << resx1 << " " << resy1 << " " << resx2 << " " << resy2 << endl;
    }
    return 0;
}

C. Challenging Cliffs

题意： 给定一个长度为n的数组a，要求将数组a重新排列。使得保证 $|a_1-a_n|$ 最小的情况下，使得$a_i<a_{i+1}$ 的个数最少。$1<=t<=100,2<=n<=2\ast 10^5,1<=a_i<=10^9$

题解： 贪心。将数组 $a$ 从小到大排序，然后找到距离最近的两个 $a_i$ 和 $a_{i+1}$ ，然后按照 $a_{i+1},a_{i+2},...,a_n,a_1,a_2,...,a_i$ 排序，这样 $a_i<a_{i+1}$ 的个数为 $n-2$

代码：

#include <bits/stdc++.h>

#define int long long
#define debug(x) cout << #x << " = " << x << endl;
using namespace std;

inline int read() {
    
   int s = 0, w = 1;
   char ch = getchar();
   while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1; ch = getchar();}
   while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
   return s * w;
}

int const MAXN = 2e5 + 10;
int n, m, T;
int a[MAXN];

signed main() {
    
    cin >> T;
    while(T--) {
    
        cin >> n;
        for (int i = 1; i <= n; ++i) cin >> a[i];
        sort(a + 1, a + 1 + n);
        if (n == 2) {
    
            for (int i = 1; i <= n; ++i) cout << a[i] << " ";
            continue;
        }
        int Min = 1e18, idx = 1;
        for (int i = 1; i <= n - 1; ++i) {
    
            if (a[i + 1] - a[i] < Min) {
    
                Min = a[i + 1] - a[i];
                idx = i;
            }
        }
        for (int i = idx + 1; i <= n; ++i) cout << a[i] << " ";
        for (int i = 1; i <= idx; ++i) cout << a[i] << " ";
        puts("");
    }
    return 0;
}

D. Deleting Divisors

题意： A和B玩博弈论游戏，A先手。每次操作允许将当前数字n减去它的一个因子d，但是要求$d\ne 1\&\&d\ne n$。不能操作的人输。现在给定n，问你先手必胜还是后手必胜。$1<=t<=10^4,1<=n<=10^9$

题解： 博弈+分类讨论。

1. 如果n为偶数：

   1. n不是 $2^k$ ，那么n减掉一个因子后，变为奇数，最后一定是1或者一个质因子，那么该状态为必胜态。
   2. n是 $2^k$ ，那么如果减掉一个因子，仍然是偶数。如果变为非 $2^k$， 则对手必胜，因此必然减去 $2^{k-1}$ ，那么变为 $2^{k-1}$。这样的话，如果k为奇数，那么先手必败；如果k为偶数，那么先手必胜。

2. 如果n为奇数：

   1. 如果n为1或者质数，必败态
   2. 如果n为其他数字，那么减去一个因子后，必然转换为非 $2^k$ 型的偶数，那么对手变为必胜态，则当前状态为必败态

   因此，如果n为奇数，为必败态

代码：

#include <bits/stdc++.h>

#define int long long
#define debug(x) cout << #x << " = " << x << endl;
using namespace std;

inline int read() {
    
   int s = 0, w = 1;
   char ch = getchar();
   while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1; ch = getchar();}
   while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
   return s * w;
}

int const MAXN = 2e5 + 10;
int n, m, T;

signed main() {
    
    cin >> T;
    while(T--) {
    
        cin >> n;
        if (n & 1) {
    
            cout << "Bob\n";
            continue;
        }
        int cnt = 0;
        while(n % 2 == 0) {
    
            n /= 2;
            cnt++;
        }
        if (n == 1) {
    
            if (cnt & 1) cout << "Bob\n";
            else cout << "Alice\n";
        }
        else {
    
            cout << "Alice\n";
        }
    }
    return 0;
}

E1. Erase and Extend (Easy Version)

题意： 给定一个长度为n的字符串s，每次操作可以删去s的最后一个字符，或者s = s + s。问如何操作使得当s长度为m时的字典序最小，打印这个最小字典序的字符串。$1<=n,k<=5000$

题解： 结论+贪心。有个结论：最后字符串必然是s的某个前缀重复若干次得到，因此直接枚举s的前缀，然后暴力得到所有长度为m的字符串，比较字典序即可。

代码：

#include <bits/stdc++.h>

#define int long long
#define debug(x) cout << #x << " = " << x << endl;
using namespace std;

inline int rd() {
    
   int s = 0, w = 1;
   char ch = getchar();
   while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1; ch = getchar();}
   while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
   return s * w;
}

int const MAXN = 2e5 + 10;
int n, m, T;
string s;
string ss[MAXN];

signed main() {
    
    cin >> n >> m;
    cin >> s;
    string pre = "";
    for (int i = 1; i <= n; ++i) {
    
        pre += s[i - 1];
        ss[i] = "";
        int t = (int)ceil((double)m / (double)i);
        for (int j = 1; j <= t; ++j) ss[i] += pre;
        ss[i].resize(m);
        // cout << ss[i] << endl;
    }
    sort(ss + 1, ss + 1 + n);
    cout << ss[1];
    return 0;
}