A guess the maximum

问题：

翻译一下就是求所有相邻元素中max - 1的最小值

代码：

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 5e4;

int a[N];
int n;

void solve() {
    cin >> n;
    int ans = 0x3f3f3f3f;
    for(int i = 1; i <= n; i ++ ) cin >> a[i];
    for(int i = 1; i <= n - 1; i ++ ) {
        int k = max(a[i], a[i + 1]) - 1;
        ans = min(ans, k);
    }
    cout << ans << endl;
}

int main() {
    int t;
    cin >> t;
    while(t -- ) {
        solve();
    }
    return 0;
}

B Xor sequences

题目：

思路：guess题，就是求两个数的lowbit

代码：

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 5e4;

int a[N];
int n;

void solve() {
    int x, y;
    cin >> x >> y;
    int ans = 1;
    for(int i = 0; i <= 30; i ++ ) {
        int xx = x >> i & 1;
        int yy = y >> i & 1;
        if(xx == yy) ans *= 2;
        else break;
    }
    cout << ans << endl;
}

int main() {
    int t;
    cin >> t;
    while(t -- ) {
        solve();
    }
    return 0;
}

C earning on bets

题目：

思路：

以下为不严谨证明

假设有n = 3 

k1 k2 k3

x1 x2 x3

满足x1 * k1 > x1 + x2 + x3

        x2 * k2 > x1 + x2 + x3

        x3 * k3 > x1 + x2 + x3

变形后有sigma 1/k < 1

由于分数的精度不好计算，因此考虑乘上所有k的lcm

即 lcm * sigma 1/k < lcm

这个是判断条件，如果成立给每个1/k乘上lcm即可

代码：

#include <iostream>

using namespace std;

const int N = 2e5 + 10;

int k[N];

int _gcd(int a, int b) {
    return b? _gcd(b, a % b): a;
}

int lcm(int a, int b) {
    return a * b / _gcd(a, b);
}

void solve() {
    int n;
    cin >> n;
    for(int i = 1; i <= n; i ++ ) cin >> k[i];
    int L = 1;
    for(int i = 1; i <= n; i ++ ) {
        L = lcm(L, k[i]);
    } 

    int sum = 0;
    for(int i = 1; i <= n; i ++ ) {
        sum += L / k[i];
    }

    if(sum >= L) cout << -1;
    else for(int i = 1; i <= n; i ++ ) {
        cout << L / k[i] << " ";
    }
}

int main() {
    int t;
    cin >> t;
    while( t-- ) {
        solve();
    }
}

D fixing a binary string

题目：

思路：对原操作进行化简，实际上就是把前p个字符翻转，并且接到原字符串的后面，则此时我们的答案串实际上是已知的如果第p个字符是1，那么答案串就是以1结尾的k pop串，反之则是以0

结尾的k pop串，既然答案已知我们遍考虑枚举p，之后在o1内用哈希字符串比较。注意到还要对原串翻转，因此还要倒着求一遍哈希

这题没有卡哈希

代码：

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 2e5 + 10;
const int P = 131;

typedef unsigned long long ULL;

int n, k;
char s[N], str[N];
ULL ansh[N], h[N], p[N], reh[N];
/*
re (n - p + 1 , n) == ans (n - p + 1, n);
h p + 1, n == ans 1, n - p
*/
bool check(int x) {
    if(reh[n] - reh[n - x] * p[x] == ansh[n] - ansh[n - x] * p[x]) {
        if(h[n] - h[x] * p[n - x] == ansh[n - x] - ansh[0] * p[n - x]) {
            return true;
        }       
    }
    return false;
}

void solve() {
    cin >> n >> k;
    for(int i = 1; i <= n; i ++ ) cin >> s[i];
    p[0] =  1;
    int c = s[1] - '0';
    for(int i = n, cnt = 1, judge = 1; i; i --, judge ++ ) {
        if(cnt & 1) str[i] = c + '0';
        else str[i] = !c + '0';
        if(judge % k == 0) cnt ++;
    }
    
    for(int i = 1; i <= n; i ++ ) p[i] = p[i - 1] * P;
    for(int i = 1; i <= n; i ++ ) h[i] = h[i - 1] * P + s[i];
    for(int i = 1; i <= n; i ++ ) ansh[i] = ansh[i - 1] * P + str[i];
    reverse(s + 1, s + n + 1);
    for(int i = 1; i <= n; i ++ ) reh[i] = reh[i - 1] * P + s[i];
    
    for(int i = 1; i <= n; i ++ ) {
        if(check(i)) {
            cout << i << endl;
            return;
        }
    }
    cout << -1 << endl;
}

int main() {
    int t;
    cin >> t;
    while(t -- ) {
        solve();
    }
    return 0;
}