MSU Trinity Contest Petrozavodsk Winter Camp 2014

题目列表

• A.MEX-Query
• D. Short Enough Task
• F. Just Another Sequence Problem

A.ABBA

题意：区间查询mex

Solution

思路:权值线段树，离线

Code

const int N = 1000010;
int n, q, a[N], ans[N];
vector> qu[N];
struct node {
	int val;
} seg[N * 4];
void update(int id) {
	seg[id].val = min(seg[id * 2].val, seg[id * 2 + 1].val);
}
void change(int id, int l, int r, int pos, int val) {
	if (l == r) {
		seg[id].val = val;
	} else {
		int mid = (l + r) / 2;
		if (pos <= mid) change(id * 2, l, mid, pos, val);
		else change(id * 2 + 1, mid + 1, r, pos, val);
		update(id);
	}
}
int search(int id, int l, int r, int d) {
	if (l == r) return l;
	int mid = (l + r) / 2;
	if (seg[id * 2].val < d) return search(id * 2, l, mid, d);
	else return search(id * 2 + 1, mid + 1, r, d);
} 
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
		a[i] = min(a[i], n + 1);
	}
	scanf("%d", &q);
	for (int i = 1; i <= q;i++) {
		int l, r;
		scanf("%d%d", &l, &r);
		qu[r].push_back({l, i});
	}
	for (int r = 1; r <= n; r++) {
		change(1, 0, n + 1, a[r], r);
		for (auto que : qu[r]) {
			ans[que.second] = search(1, 0, n + 1, que.first);
		}
	}
	for (int i = 1; i <= q; i++) {
		printf("%d\n", ans[i]);
	}
}

D. Short Enough Task

略

Solution

思路:枚举回文串长度即可，那么贡献累加就是就是$k^{(i/2)}*(n-i+1)/k^i$，由于越往后误差越大，枚举到一定次数就够了

Code

void solve() {
    ll n, k;
    cin >> n >> k;
    long double ans = n;
    if (k == 1) {
        cout << fixed << setprecision(10) << (long double)(1 + n)*(long double)n/ 2.0 << endl;
    } else {
        long double   tmp = 1;
        for (int i = 1; i <= n && i <= 1e6; ++i) {
            if((i&1)){
                tmp*=k;
            }
            ans += ( ((n - i) * 1.0) / tmp); 
        }
        cout << fixed << setprecision(10) << ans << endl;
    }
}

F.Just Another Sequence Problem

题意：略

Solution

思路:设$dp[i][j]$表示当前以$i$结尾以$j$开头的最小代价，$O(n^3)$被卡过去了，正解应该是斜率优化

Code

const int N = 2010;
/* dp[i][j]代表最后一块是i->j*/
dp[i][j] = dp[k][i] 
ll dp[N][N];
void solve() {
    int n;
    cin >> n;
    vector a(n + 1), pre(n + 1);
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
        pre[i] = pre[i - 1] + a[i];
    }
    for (int i = 2; i <= n; ++i) {
        for (int j = i; j <= n; ++j) {
            dp[i][j]=INT_MIN;
            for (int k = 1; k <= i - 1; ++k) {
                dp[i][j] = max(dp[i][j], dp[k][i - 1] + (pre[i - 1] - pre[k - 1]) * (pre[j] - pre[i - 1]));
            }
        }
    }
    ll ans =0;
    for (int i = 1; i <= n; ++i) {
        ans = max(ans, dp[i][n]);
    }
    cout << ans << '\n';
}