Codeforces Round #667 (Div. 3)

A

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
#define IOS ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef double db;
 
const int N = 1e5 + 5;
 
int n, m, _, k;
 
int main() {
    IOS;
    for (cin >> _; _; --_) {
        cin >> n >> m;
        n = abs(n - m);
        if(!n) cout << 0 << '\n';
        else cout << ((n - 1) / 10 + 1) << '\n';
    }
    return 0;
}

B

设个函数求导, 发现减的越多越好

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
#define IOS ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef double db;
 
const int N = 1e5 + 5;
 
int n, m, _, k;
 
int main() {
    IOS;
    for (cin >> _; _; --_) {
        ll a, b, x, y, n; cin >> a >> b >> x >> y >> n;
        ll c = min(n, a - x);
        ll a1 = a - c, b1 = max(b - n + c, y);
        c = min(n, b - y);
        ll b2 = b - c, a2 = max(a - n + c, x);
        if (a1 * b1 < a2 * b2) cout << a1 * b1 << '\n';
        else cout << b2 * a2 << '\n'; 
    }
    return 0;
}

C

这里是薄弱, 花太多时间了

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
#define IOS ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef double db;
 
const int N = 1e5 + 5;
 
int n, m, _, k;
int x, y, z, ans, xd, xa;
 
void check(int d) {
    int c = z / d, e = x % d;
    if (!e) e = d;
    if (c >= n) return;
    if ((y - e) / d >= n)
        e = y - (n - 1) * d;
 
    if (e + (n - 1) * d < ans) {
        ans = (n - 1) * d + e;
        xa = e, xd = d;
    }
}
 
int main() {
    IOS;
    for (cin >> _; _; --_) {
        cin >> n >> x >> y;
        z = y - x; ans = 1e9;
        rep (i, 1, z / i) 
            if (z % i == 0) {
                check(i);
                if (i * i != z) check(z / i);
            }
 
        xa -= xd;
        rep (i, 1, n) cout << xa + i * xd << ' ';
        cout << endl;
    }
    return 0;
}

D

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
#define IOS ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef double db;
 
const int N = 1e5 + 5;
 
int n, m, _, k;
int a[20], tot;
ll s[20] = {1};
 
int main() {
    IOS;
    rep (i, 1, 18) s[i] = s[i - 1] * 10;
    for (cin >> _; _; --_) {
        ll n, ans = 0; cin >> n >> m; k = tot = 0;
        while (n) {
            a[tot++] = n % 10; n /= 10;
            k += a[tot - 1];
        }
        for (int i = 0; i < tot && k > m; ++i) {
            if (a[i]) {
                k -= a[i]; ans += s[i] * (10 - a[i]);
                for (int j = i + 1; ; ++j) {
                    if (j == tot) { a[tot++] = 1; break; }
                    else if (a[j] == 9) a[j] = 0, k -= 9;
                    else { ++a[j], ++k; break; }
                }
            }
        }
        cout << ans << endl;
    }
    return 0;
}

E

a[i] 表示从(排好序) 第 i 个点放木板可接住的点的数量, b[i] 表示 在 i~n max(a[i]), 枚举就行

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
#define IOS ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef double db;
 
const int N = 2e5 + 5;
 
int n, m, _, k;
PII p[N];
int a[N], b[N], c[N];
 
int main() {
    IOS;
    for (cin >> _; _; --_) {
        cin >> n >> k; m = b[n + 1] = 0;
        rep (i, 1, n) cin >> p[i].fi;
        rep (i, 1, n) cin >> p[i].se;
        sort(p + 1, p + 1 + n);
        int l = 1, r = 2;
        for ( ; r <= n; ++r)
            while (p[r].fi - p[l].fi > k) a[l] = r - l, c[l++] = r;
        while (l < r) a[l] = r - l, c[l++] = r;
        per (i, n, 1) b[i] = max(b[i + 1], a[i]);
        rep (i, 1, n) m = max(m, a[i] + b[c[i]]);
        cout << m << endl;
    }
    return 0;
}

F

dp

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
#define IOS ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef double db;

const int N = 2e2 + 5;

int n, m, _, k;
int f[N][N][N];
char s[N], t[5];

int main() {
    IOS; cin >> n >> k >> s + 1 >> t + 1;
    memset(f, -0x3f, sizeof f); f[0][0][0] = 0;
    rep (i, 1, n)
        rep (j, 0, i)
            rep (x, 0, min(i, k)) 
                rep (ch, 'a', 'z') {
                    int nx = (ch == s[i] ? x : x + 1);
                    int nj = (ch == t[1] ? j + 1 : j);
                    f[i][nj][nx] = max(f[i][nj][nx], f[i - 1][j][x] + (ch == t[2] ? j : 0)); 
                }
    int ans = 0;
    rep (i, 0, n)
        rep (j, 0, k) ans = max(ans, f[n][i][j]);
    cout << ans;
    return 0;
}