Codeforces Round #413(Div. 1 + Div. 2)

---

C. Fountains

首先特判买一个C一个D的情况。
考虑买两个C(或D)，先按照价格排个序，枚举一个位置，则另一个可行位置构成一个区间。然后用ST表区间最值即可。

#include <bits/stdc++.h>using namespace std;const int MAXN = 100005;struct ft {    int b, c;    friend bool operator < (const ft &a, const ft &b)    { return a.c < b.c; }} coin[MAXN], dm[MAXN];int top1 = 0, top2 = 0;int max_c[MAXN], max_d[MAXN];int n, c, d;int f[MAXN][25], lg[MAXN];int max_val(int i, int j){    if (i > j) return 0;    int k = lg[j-i+1];    return max(f[i][k], f[j-(1<<k)+1][k]);}int max_val_without(int i, int j){    if (i < j) return max_val(1, i);    return max(max_val(1, j-1), max_val(j+1, i));}int solve(ft a[MAXN], int top, int M){    sort(a+1, a+top+1);    memset(f, 0, sizeof f);    for (int i = 1; i <= top; i++) f[i][0] = a[i].b;    for (int j = 1; j <= 20; j++)         for (int i = 1; i <= top; i++) {            f[i][j] = f[i][j-1];            if (i+(1<<(j-1)) <= top) f[i][j] = max(f[i][j], f[i+(1<<(j-1))][j-1]);        }    int Rt = top, ans = 0;    for (int i = 1; i <= top; i++) {        if (a[i].c > M) break;        while (Rt && (a[Rt].c+a[i].c > M || Rt == i)) Rt--;        if (Rt == 0) break;        ans = max(ans, max_val_without(Rt, i)+a[i].b);    }     //cout << ans << endl;    return ans;}int main(){    scanf("%d %d %d", &n, &c, &d);    int j = 1, t = 0;    for (int i = 1; i <= n; i++) {        while (j*2 <= i) j <<= 1, t++;        lg[i] = t;        //cout << lg[i] << " ";    }    //cout << endl;    for (int i = 1; i <= n; i++) {        int a, b; char ch;        scanf("%d %d %c", &a, &b, &ch);        // cout << a << " " << b << "," << c << endl;        if (ch == 'C') coin[++top1].b = a, coin[top1].c = b, max_c[b] = max(max_c[b], a);        else dm[++top2].b = a, dm[top2].c = b, max_d[b] = max(max_d[b], a);    }    max_d[0] = 0, max_c[0] = 0;    for (int i = 1; i <= c; i++) max_c[i] = max(max_c[i], max_c[i-1]);    for (int i = 1; i <= d; i++) max_d[i] = max(max_d[i], max_d[i-1]);    // case1     int ans = 0;    if (max_d[d] != 0 && max_c[c] != 0)        ans = max(ans, max_d[d]+max_c[c]);    // case2    ans = max(ans, max(solve(coin, top1, c), solve(dm, top2, d)));    cout << ans << endl;    return 0;}

D. Field expansion

大力爆搜..CF机子也真是快…

#include <bits/stdc++.h>using namespace std;long long a, b, h, w, n;long long ai[100005];long long dat[100005], top = 0;int tms[100005];int dfs(int nd, long long h, long long w){    //cout << dat[nd] << ' ' << h << ' ' << w << endl;    if ((h >= a && w >= b) ||(h >= b && w >= a)) return 0;    if (nd == 0) return 2333333;    long long th = 1, tw = 1;    int ans = 2333333;    for (int i = 0; i <= tms[dat[nd]]; i++) {        tw = 1;        for (int j = 0; j+i <= tms[dat[nd]]; j++) {            ans = min(ans, dfs(nd-1, h*th, w*tw)+i+j);            if (w*tw >= 100000) break;            tw *= dat[nd];        }        if (h*th >= 100000) break;        th *= dat[nd];    }    return ans;}int main(){    scanf("%I64d%I64d%I64d%I64d%I64d", &a, &b, &h, &w, &n);    for (int i = 1; i <= n; i++) scanf("%I64d", &ai[i]);    sort(ai+1, ai+n+1);     for (int i = n; i >= max(1ll, n-40); i--) {         tms[ai[i]]++;        if (dat[top] != ai[i])            dat[++top] = ai[i];        // cout << i << endl;    }    int ans = dfs(top, h, w);    if (ans < 2333333)    cout << dfs(top, h, w) << endl;    else puts("-1");    return 0;}