Codeforces Round 870 (A-D)

Posted 2023-05-06 huaiii

A

\\(n^2\\)暴力推，遍历可能撒谎的人数(0-n)，然后\\(O(n)\\)check就行了。
仔细看逻辑其实\\(O(1)\\)就能check，草率了，后面再修正

点击查看代码

#include <bits/stdc++.h>
#define rep(i, l, r) for(int i = l; i <= r; i++)
#define per(i, r, l) for(int i = r; i >= l; i--)
#define debug(tar, l, r) rep(ii, l, r) cout << tar[ii] << " \\n"[ii==r]

using namespace std;
using ll = long long;
using pii = pair<int,int>;
const int N = 1e5+5;
const int mod = 1e9+7;

int n, a[N];

bool chk(int x)
    rep(i, 1, n-x)
        if(a[i] > x) return false;
    
    rep(i, n-x+1, n)
        if(a[i] <= x) return false;
    
    return true;


void solve()
    cin >> n;
    rep(i, 1, n) cin >> a[i];
    sort(a+1, a+1+n);
    rep(i, 0, n)
        if(chk(i))
            cout << i << "\\n";
            return ;
        
    
    cout << "-1\\n";

int main()
    std::ios::sync_with_stdio(false);
    int tt; cin >> tt; while(tt--)
        solve();
    return 0^0;

B

回文对应的两个数的差值一定是\\(x\\)的倍数，把所有情况取个gcd即可

点击查看代码

#include <bits/stdc++.h>
#define rep(i, l, r) for(int i = l; i <= r; i++)
#define per(i, r, l) for(int i = r; i >= l; i--)
#define debug(tar, l, r) rep(ii, l, r) cout << tar[ii] << " \\n"[ii==r]

using namespace std;
using ll = long long;
using pii = pair<int,int>;
const int N = 1e5+5;
const int mod = 1e9+7;

int n, b[N];

void solve()
    cin>> n;
    rep(i, 1, n) cin >> b[i];
    int ans = 0;
    rep(i, 1, n/2)
        int dif = abs(b[i]-b[n-i+1]);   
        ans = __gcd(ans, dif);
    
    cout << ans << "\\n";
   
int main()
    std::ios::sync_with_stdio(false);
    int tt; cin >> tt; while(tt--)
        solve();
    return 0^0;

C

n除1外的最小的因子小于等于m即可

点击查看代码

#include <bits/stdc++.h>
#define rep(i, l, r) for(int i = l; i <= r; i++)
#define per(i, r, l) for(int i = r; i >= l; i--)
#define debug(tar, l, r) rep(ii, l, r) cout << tar[ii] << " \\n"[ii==r]

using namespace std;
using ll = long long;
using pii = pair<int,int>;
const int N = 1e6+5;
const int mod = 1e9+7;

int n, m;
bool no[N];
vector <int> p;
void init()
    rep(i, 2, 1000000)
        if(!no[i]) p.push_back(i);
        for(auto& pp:p)
            if(pp*i > 1000000) break;
            no[pp*i] = 1;
        
    


void solve()
    cin >> n >> m;
    if(n > 1 && n <= m)
        cout << "NO\\n";
        return ;
    
    for(auto& pp:p)
        if(pp*pp > n) break;
        if(n%pp == 0)
            if(pp <= m)
                cout << "NO\\n";
                return ;
            
        
    
    cout << "YES\\n";

int main()
    std::ios::sync_with_stdio(false);
    init();
    int tt; cin >> tt; while(tt--)
        solve();
    return 0^0;

D

做法是尺取，不知道会不会假了
将\\(b_i1+b_i2+b_i3-(r-l)\\)转换成\\((b_l+l)+(b_r-r)+b_i\\)，其中\\(l < i < r\\)
对\\(b_i+i\\)求前缀最值，对\\(b_i-i\\)求后缀最值，中间\\(b_i\\)用st表，然后尺取

点击查看代码

#include <bits/stdc++.h>
#define rep(i, l, r) for(int i = l; i <= r; i++)
#define per(i, r, l) for(int i = r; i >= l; i--)
#define debug(tar, l, r) rep(ii, l, r) cout << tar[ii] << " \\n"[ii==r]

using namespace std;
using ll = long long;
using pii = pair<int,int>;
const int N = 1e5+5;
const int mod = 1e9+7;

int n, b[N], p[N], s[N], f[N][30];
void pre()
    rep(i, 1, n) f[i][0] = b[i];
    for (int k = 1; (1 << k) <= n; ++k)
        for (int i = 1; i + (1 << k) - 1 <= n; ++i)
            f[i][k] = max(f[i][k - 1], f[i + (1 << (k - 1))][k - 1]);

int ask(int l, int r)
    int k = log2(r - l + 1);
    return max(f[l][k], f[r - (1 << k) + 1][k]);



void solve()
    cin >> n;
    rep(i, 1, n) cin >> b[i];
    pre();
    p[0] = s[n+1] = -1e9;
    rep(i, 1, n) p[i] = max(p[i-1], b[i]+i);
    per(i, n, 1) s[i] = max(s[i+1], b[i]-i);
    int l = 1;
    int ans = 0;
    for(int i = 3; i <= n; i++)
        while(l+1 <= i-2 && p[l+1]+ask(l+2, i-1) >= p[l]+ask(l+1, i-1)) l++;
        ans = max(ans, p[l]+ask(l+1, i-1)+s[i]);
    
    cout << ans << "\\n";

int main()
    std::ios::sync_with_stdio(false);
    int tt; cin >> tt; while(tt--)
        solve();
    return 0^0;

Codeforces Edu Round 59 A-D

A. Digits Sequence Dividing

注意特殊情况，在\(n = 2\)时除非\(str[1] >= str[2]\)，否则可以把第一个数划分下来，剩下的数直接当成一组，一定满足条件。

#include <cstdio>
#include <iostream>
#include <string>
using namespace std;
const int N = 310;
int n;
char s[N];
int main()
    int T; scanf("%d", &T);
    while(T--)
        scanf("%d%s", &n, s + 1);
        if(n == 2)
            if(s[1] >= s[2]) puts("NO");
            else printf("YES\n2\n%c  %c\n", s[1], s[2]);
        else

            printf("YES\n2\n%c ", s[1]);
            for(int i = 2; i <= n; i++) putchar(s[i]);
            puts("");
        
    
    return 0;

B. Digital root

通过打表找规律发现的…看了题解，证明还是很NB的...

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long LL;
int main()
    int T; scanf("%d", &T);
    while(T--)
        LL k, x; scanf("%lld%lld", &k, &x);
        printf("%lld\n", (k - 1) * 9 + x);
    
    return 0;

C. Brutality

用堆维护连续子段最大和即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long LL;
const int N = 200010;
int n, k, a[N];
LL ans = 0;
char s[N];
priority_queue<int, vector<int>, greater<int> > q;
int main()
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; i++) scanf("%d", a + i);
    scanf("%s", s + 1);
    for(int i = 1; i <= n; i++)
        if(s[i] != s[i - 1])
            while(!q.empty()) ans += q.top(), q.pop();
            q.push(a[i]);
        else
            q.push(a[i]);
        
        while(q.size() > k) q.pop();
    
    while(!q.empty()) ans += q.top(), q.pop();
    printf("%lld\n", ans);
    return 0;

D. Compression

实质上是把这图压缩到最小的点阵图，用\(bitset\)优化复杂度，暴力水过。

#include <cstdio>
#include <iostream>
#include <bitset>
using namespace std;
const int N = 5210;
bitset<N> a[N];
int n; 
//密集恐惧症
int main()
    ios::sync_with_stdio(false);
    cin >> n;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j += 4)
            char x; cin >> x;
            //16进制转2进制
            if(x == '1') a[i][j + 3] = 1;
            else if(x == '2') a[i][j + 2] = 1;
            else if(x == '3') a[i][j + 2] = a[i][j + 3] = 1;
            else if(x == '4') a[i][j + 1] = 1;
            else if(x == '5') a[i][j + 1] = a[i][j + 3] = 1;
            else if(x == '6') a[i][j + 2] = a[i][j + 3] = 1;
            else if(x == '7') a[i][j + 1] = a[i][j + 2] = a[i][j + 3] = 1;
            else if(x == '8') a[i][j] = 1;
            else if(x == '9') a[i][j] = a[i][j + 3] = 1;
            else if(x == 'A') a[i][j] = a[i][j + 2] = 1;
            else if(x == 'B') a[i][j] = a[i][j + 2] = a[i][j + 3] = 1;
            else if(x == 'C') a[i][j] = a[i][j + 1] = 1;
            else if(x == 'D') a[i][j] = a[i][j + 1] = a[i][j + 3] = 1;
            else if(x == 'E') a[i][j] = a[i][j + 1] = a[i][j + 2] = 1;
            else if(x == 'F') a[i][j] = a[i][j + 1] = a[i][j + 2] = a[i][j + 3] = 1;
        
    
    for(int x = n; x >= 2; x--)
        if(n % x) continue;
        bool ep = true;
        for(int i = 1; i <= n; i += x)
            for(int j = i + 1; j < i + x; j++)
                ep = ep && (a[j] == a[i]);
                if(!ep) break;
            
            if(!ep) break;
            for(int j = 1; j <= n; j += x)
                for(int k = j + 1; k < j + x; k++)
                    ep = ep && (a[i][k] == a[i][k - 1]);
                    if(!ep) break;
                
                if(!ep) break;
            
            if(!ep) break;
        
        if(ep)  cout << x; return 0; 
    
    cout << 1;
    return 0;