Codeforces Round #345 (Div. 2)

DFS A-joysticks

Too troublesome direct Dfs burst search bar, there is a pit point is the current power <=1 can no longer power down, the direct end.

#include <bits/stdc++.h>typedef long long ll;const int N = 1e5 + 5;int ans = 0;void DFS (int A, int b, int step) {
   if (a <= 0 | | b <= 0) {        ans = std::max (ans, step); return;    }    if (A < b && b-2 >= 0) {        DFS (A + 1, b-2, step + 1);    } else if (a >= b && a-2 >= 0) {        DFS (a-2, B + 1, step + 1);}    } int main () {    int A, b;   Std::cin >> a >> b;    Ans = 0;    DFS (A, b, 0);    Std::cout << ans << ' \ n ';    return 0;}

Construction + greedy b-beautiful paintings

Every time you remove an increment sequence that is not repeated until the collection is empty

#include <bits/stdc++.h>typedef long long ll;const int n = 1e5 + 5;std::vector<int> vec;int main () {    int n ;  Std::cin >> N;    for (int x, i=0; i<n; ++i) {        std::cin >> x; vec.push_back (x);    }    Std::sort (Vec.begin (), Vec.end ());    int ans = 0;    while (Vec.size () > 0) {        std::vector<int> tmp;        int pre = 0, num = 0;        for (auto X:vec) {            if (x > Pre) {                num++; pre = x;            } else {                tmp.push_back (x);            }        }        Ans + = num-1; VEC = tmp;    }    Std::cout << ans << ' \ n ';    return 0;}

Math + tolerance C-watchmen

Simplification formula gets found pair (i, j) Xi = = XJ | | Yi = = yj. Find two times and let it go. Map to do better.

#include <bits/stdc++.h>typedef long long ll;const int N = 2e5 + 5;const int INF = 1e9 + 7;std::p air<int, int> Point[n];ll Calc (int N) {return 1LL * N * (n-1)/2;} BOOL CMPX (std::p air<int, int> A, std::p air<int, int> b) {return A.first < B.first;} BOOL Cmpy (std::p air<int, int> A, std::p air<int, int> b) {return A.second < B.second;}    int main () {int n; scanf ("%d", &n);        for (int x, y, i=0; i<n; ++i) {scanf ("%d%d", &x, &y);    Point[i] = Std::make_pair (x, y);    } std::sort (Point, Point+n, cmpx);    ll ans = 0;    int x = inf, y = inf, num = 1;        for (int i=0; i<n; ++i) {if (x = = Point[i].first) {num++;            } else {x = Point[i].first;            if (num > 1) {ans + = calc (num);        num = 1;    }} if (num > 1) {ans + = calc (num);    } std::sort (Point, Point+n, cmpy); x = INF; y = INF; num = 1;        for (int i=0; i<n; ++i) {if (y = = Point[i].second) {num++;            } else {y = Point[i].second;            if (num > 1) {ans + = calc (num);        num = 1;    }} if (num > 1) {ans + = calc (num);    } std::sort (point, Point+n); x = INF; y = INF;    num = 1;        for (int i=0; i<n; ++i) {if (x = = Point[i].first && y = = Point[i].second) {num++;            } else {x = Point[i].first, y = point[i].second;            if (num > 1) {ans-= calc (num);        num = 1;    }} if (num > 1) {ans-= calc (num);    } printf ("%i64d\n", ans); return 0;}

Points D-image Preview

Test instructions: Browse pictures, browse, swipe, and invert takes time to ask how many images to browse in T time.

Analysis: Define two pointers from,to, the feasible scheme is n+1->from,from->n+1,n+1->to or N+1->to,to->n+1,n+1->from, and from->to. The first two repeat sliding can choose a small distance, the third as long as the definition of to=n+1 is right. In fact, you can do it with two points.

#include <bits/stdc++.h>typedef long long ll;const int n = 5e5 + 5;char str[2*n];ll tim[2*n];int N, a, B, T;ll get_t IME (int from, int. to) {    LL ret = Tim[from]-tim[to-1];    int move = from-to + std::min (from-(n+1), (n+1)-to);    return ret + 1ll * A * move;} int main () {    scanf ("%d%d%d%d", &n, &a, &b, &t);    scanf ("%s", str + 1);    for (int i=1; i<=n; ++i) {        Tim[i] = (str[i] = = ' W '? (b+1): 1);    }    for (int i=n+1; i<=2*n; ++i) {        tim[i] = tim[i-n];    }    for (int i=1; i<=2*n; ++i) {        Tim[i] + = tim[i-1];    }    int ans = 0;    int from = n + 1;    for (int to=2; to<=n+1; ++to) {while        (from < to+n-1 && Get_time (from+1, to) <= T) from++;        if (Get_time (from, to) <= T) ans = std::max (ans, from-to + 1);    }    printf ("%d\n", ans);    return 0;}

E-table Compression

Test instructions: To a matrix, the new matrix, so that the original matrix peers, the same column size relationship is not changed, and make the maximum value of the novel matrix is minimal. i.e. discretization

Analysis:

Codeforces Round #345 (Div. 2)