平均數

題意是讓求平均值第K小的連續子區間。
發現直接計算無論怎麼優化都是 \(n^2\) 的，然後發現這樣找K個的似乎可以考慮二分答案。
簡單推一下式子。
記 \(sum[i]\) 為字首和，顯然符合條件的區間有：

即

預設 \(j < i\)，區間為 \([i+1,j]\)，所以顯然下標從0開始。

所以問題轉化為一個類似01分數規劃的東西，check的話用樹狀陣列求逆序對，注意處理下標為0。

程式碼寫得有點長，而且沒有卡常所以有點慢。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn = 1e5 + 10;
const double eps = 1e-8;
double sum[maxn];
int w[maxn];
long long n, k;
void Add(int i, int val) {
    for (; i <= 100001; i += (i & (-i))) w[i] += val;
}
int Ask(int i) {
    int ans = 0;
    for (; i; i -= i & (-i)) ans += w[i];
    return ans;
}
struct node {
    double data;
    int pos;
}tmp[maxn];
bool cmp(node x, node y) { return x.data < y.data; }
bool Check(double mid) {
    long long cnt = 0;
    tmp[0] = (node) { 0, 0 };
    for (int i = 1; i <= n; i++)
	tmp[i] = (node) { sum[i] - mid * i, i};
    sort(tmp, tmp + 1 + n, cmp);
    for (int i = n; i >= 0; i--) {
	cnt += Ask(tmp[i].pos + 1);
	Add(tmp[i].pos + 1, 1);
    }
    for (int i = 0; i <= n; i++) Add(i + 1, -1);
    return cnt >= k;
}
int main() {
    freopen("ave.in", "r", stdin);
    freopen("ave.out", "w", stdout);
    scanf("%lld %lld", &n, &k);
    for (int i = 1; i <= n; i++) {
	scanf("%lf", &sum[i]);
	sum[i] += sum[i - 1];
    }
    double l = 1.0, r = 1000000000.0, mid;
    while (r - l > eps) {
	mid = (l + r) / 2;
	//cout<<mid<<endl;
	if (Check(mid)) r = mid;
	else l = mid;
    }
    printf("%.4lf\n", l);
    return 0;
}
 

塗色遊戲

並不是很會推這個式子...c一波學長題解吧...

設 \(f[i][j]\) 代表第\(i\)列選\(j\)個顏色的方案數，

\(g[i][j]\)代表用任意\(i\)個顏色填\(j\)個塊的方案數，

\(h[i][j]\)代表上一列選\(i\)個顏色這一列選\(j\)個的方案數

\(g[i][j]=g[i-1][j-1] \times (p-i+1)+g[i][j-1] \times i\)，就是第二類斯特林數;

\(h[j][k]=\frac{g[k][n]}{C_{p}^{k}} \times \sum_{x=max(q,j,k)}^{min(p, j+k)}C_{j}^{j+k-x}C_{p-j}^{x-j}\)

\(f[i][j]=f[i-1][k] \times h[k][j]\)

發現\(f\)的轉移跟\(i\)無關，再加上\(m\)如此巨大，可以矩陣乘。

其實並不用真正寫一個矩陣快速冪，但我還是無腦套了板子...

另外，注意卡常。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 105;
const int mo = 998244353;
int n, m, p, q;
long long g[maxn][maxn], h[maxn][maxn], c[maxn][maxn];
struct Mat {
    long long w[maxn][maxn];
    Mat() {
	memset(w, 0, sizeof w);
    }
    void init() {
	for (int i = 1; i <= n; i++) {
	    for (int j = 1; j <= n; j++) {
		w[i][j] = (i == j ? 1 : 0);
	    }
	}
    }
    Mat operator * (const Mat &b) const {
	Mat c;
	for (int i = 1; i <= n; i++) {
	    for (int j = 1; j <= n; j++) {
		for (int k = 1; k <= n; k++) {
		    c.w[i][j] += w[i][k] * b.w[k][j] % mo;
		}
		c.w[i][j] %= mo;
	    }
	}
	return c;
    }
};
long long Qpow(long long x, int t) {
    long long s = 1;
    while (t) {
	if (t & 1) s = s * x % mo;
	x = x * x % mo;
	t >>= 1;
    }
    return s;
}
Mat Matpow(Mat x, int t) {
    Mat s;
    s.init();
    while (t) {
	if (t & 1) s = s * x;
	x = x * x;
	t >>= 1;
    }
    return s;
}
int main() {
    //freopen("color.in", "r", stdin);
    //freopen("color.out", "w", stdout);
    cin >> n >> m >> p >> q;
    for (int i = 0; i <= p; i++) 
	c[i][0] = 1;
    for (int i = 1; i <= p; i++)
	for (int j = 1; j <= i; j++)
	    c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mo;
    g[0][0] = 1, g[1][1] = 1, g[2][1] = 1;
    for (int i = 1; i <= p; i++) {
	for (int j = 1; j <= n; j++) {
	    g[i][j] = (g[i - 1][j - 1] * (p - i + 1) % mo + g[i][j - 1] * i % mo) % mo;
	}
    }
    for (register int j = 1; j <= n; j++) {
	for (register int k = 1; k <= n; k++) {
	    long long sum = 0;
	    for (int x = max(q, max(j, k)); x <= min(p, j + k); x++)
		sum = (sum + c[j][j + k - x] * c[p - j][x - j] % mo) % mo;
	    h[j][k] = g[k][n] * Qpow(c[p][k], mo - 2) % mo * sum % mo;
	}
    }
    Mat base, ans;
    for (int i = 1; i <= n; i++) 
	for (int j = 1; j <= n; j++)
	    base.w[i][j] = h[i][j];
    for (int i = 1; i <= min(p, n); i++) {
	ans.w[1][i] = g[i][n];
    }
    base = Matpow(base, m - 1);
    ans = ans * base;
    long long sum = 0;
    for (int i = 1; i <= min(n, p); i++)
	sum = (sum + ans.w[1][i]) % mo;
    cout << sum << endl;
    return 0;
}

序列

考場上打了1h+的主席樹掛了...

用線段樹套vector水過。

每個節點開一個vector，存下能夠完全覆蓋此節點代表區間的詢問的x值。（注意保證vector有序）

然後考慮每個a[i]的貢獻，從根一直到區間為i的葉子節點，每到一個節點就在該節點的vector裡二分一下，由於一個詢問最多隻會更新一個節點一次，所以貢獻不會算重。

ans[i]記錄a[i]的貢獻。

修改的話直接計算更新即可。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
using namespace std;
const int maxn = 1e5 + 10;
inline int read() {
    int s = 0, w = 1;
    char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') w = -1; c = getchar(); }
    while (c >= '0' && c <= '9') s = s * 10 + c - '0', c = getchar();
    return s * w;
}
struct tree {
    vector<int> qe;
}t[maxn << 2];
#define ls (u << 1)
#define rs (u << 1 | 1)
void Update(int u, int l, int r, int ql, int qr, int val) {
    if (ql <= l && r <= qr) {
	t[u].qe.push_back(val);
	return;
    }
    int mid = (l + r) >> 1;
    if (ql <= mid) Update(ls, l, mid, ql, qr, val);
    if (qr > mid) Update(rs, mid + 1, r, ql, qr, val);
    return;
}
int Get(int u, int val) {
    int ans, flag = 1;
    /*for (int i = 0; i < t[u].qe.size(); i++) {
	if (t[u].qe[i] > val) {
	    flag = 0;
	    ans = i;
	    break;
	}
    }
    if (flag) ans = t[u].qe.size();*/
    int l = 0, r = t[u].qe.size() - 1, mid;
    while (l <= r) {
	mid = (l + r) >> 1;
	if (t[u].qe[mid] > val) r = mid - 1;
	else l = mid + 1;
    }
    //cout << "************" << endl;
    //printf("ans = %d l = %d\n", ans, l);
    //cout << "************" << endl;
    return l;
}
int Ask(int u, int l, int r, int pos, int val) {
    if (l == r) {
	return Get(u, val);
    }
    int mid = (l + r) >> 1;
    int ans = 0;
    ans += Get(u, val);
    if (pos <= mid) ans += Ask(ls, l, mid, pos, val);
    else ans += Ask(rs, mid + 1, r, pos, val);
    return ans;
}
int n, m, q;
int a[maxn];
int aa[maxn];
struct node {
    int l, r, x;
}qq[maxn];
bool cmp(node aa, node bb) { return aa.x < bb.x; }
int main() {
    freopen("seq.in", "r", stdin);
    freopen("seq.out", "w", stdout);
    n = read(), m = read(), q = read();
    for (int i = 1; i <= n; i++) a[i] = read();
    for (int i = 1; i <= m; i++) {
	qq[i].l = read(), qq[i].r = read(), qq[i].x = read();
    }
    sort(qq + 1, qq + 1 + m, cmp);
    for (int i = 1; i <= m; i++) {
	Update(1, 1, n, qq[i].l, qq[i].r, qq[i].x);
    }
    int ans = 0;
    for (int i = 1; i <= n; i++) {
	aa[i] = Ask(1, 1, n, i, a[i]);
	ans += aa[i];
    }
    printf("%d\n", ans);
    //for (int i = 1; i <= n; i++)
	//cout << aa[i] << endl;
    int u, v, pos, val;
    while (q--) {
	u = read(), v = read();
	pos = u ^ ans;
	val = v ^ ans;
	ans -= aa[pos];
	aa[pos] = Ask(1, 1, n, pos, val);
	ans += aa[pos];
	printf("%d\n", ans);
    }
}