C是真的不会, 连续区间第k大

A

取好模数就行

#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 IO ios::sync_with_stdio(0); cin.tie(0)
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, mod = 998244353;

ll n, m, _, k;
ll a, b;

int qpow(int a, int b) {
    int ans = 1;
    for (; b; b >>= 1, a = (ll)a * a % mod)
        if (b & 1) ans = (ll)ans * a % mod;
    return ans;
}

int main() {
    IO;
    int c100 = qpow(100, mod - 2);
    for (cin >> _; _; --_) {
        cin >> n >> a >> b;
        int c = __gcd(a, (ll)100);
        a = a / c * qpow(100 / c, mod - 2) % mod;
        c = __gcd(b, (ll)100);
        b = b / c * qpow(100 / c, mod - 2) % mod;
        c = __gcd(n, a);
        ll d = (n / c * qpow(a / c, mod - 2) - 1) % mod;
        m = ((n - d * b) % mod + mod) % mod;
        cout << m << '\n';
    }
    return 0;
}

B

暴力

#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 IO ios::sync_with_stdio(0); cin.tie(0)
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;
char s[N], t[N];
int w, x, y, z;
int res[N];

void work(int k, int a) {
    res[k] = w + x + 1 + x + z;
    if (!a) {
        int c = 1e9, d = 0;
        rep(i, k + 1, n) {
            if (s[i] == t[i]) ++d;
            else --d;

            if (s[i] != '1') break;
            else {
                if (t[i] == '1') c = min(d, c);
                else c = min(d + 2, c);
            }
            if (i == n) ++d;
        }
        if (k == n) ++d;
        res[k] += min(d, c);
    }
}

int main() {
    IO;
    for (cin >> _; _; --_) {
        cin >> n >> s + 1 >> t + 1;
        w = x = y = z = 0;
        rep(i, 1, n) {
            if (t[i] != s[i]) work(i, t[i] - '0');
            else res[i] = 2e5;

            w += (t[i] == '0') && (s[i] == '0');
            x += (t[i] == '1') && (s[i] == '0');
            y += (t[i] == '0') && (s[i] == '1');
            z += (t[i] == '1') && (s[i] == '1');
        }

        m = x + y, w = 0;
        per(i, n, 1) {
            m = min(m, res[i] + w);
            w += (s[i] != t[i]);
        }
        cout << m << '\n';
    }
    return 0;
}

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来源： https://www.cnblogs.com/2aptx4869/p/13468530.html