| 题号 | 标题 | 已通过代码 | 题解/讨论 | 通过率 | 团队的状态 |
|---|---|---|---|---|---|
| A | The power of Fibonacci | 点击查看 | 进入讨论 | 69/227 | 未通过 |
| B | Quadratic equation | 点击查看 | 高次剩余 | 391/888 | 未通过 |
| C | Inversions of all permutations | 点击查看 | 进入讨论 | 28/61 | 未通过 |
| D | Knapsack Cryptosystem | 点击查看 | 进入讨论 | 606/2251 | 通过 |
| E | All men are brothers | 点击查看 | 进入讨论 | 425/1117 | 通过 |
| F | Birthday Reminders | 点击查看 | 进入讨论 | 5/11 | 未通过 |
| G | Checkers | 点击查看 | 进入讨论 | 0/15 | 未通过 |
| H | Cutting Bamboos | 点击查看 | 二分，主席树 | 187/834 | 通过 |
| I | KM and M | 点击查看 | 进入讨论 | 19/296 | 未通过 |
| J | Symmetrical Painting | 点击查看 | 进入讨论 | 227/930 | 通过 |
H Cutting Bamboos
这道题用主席树过的，记录下区间权值。
// #pragma GCC optimize(2) // #pragma GCC optimize(3) // #pragma GCC optimize(4) #include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> #include <unordered_map> // #include<bits/extc++.h> // using namespace __gnu_pbds; using namespace std; #define pb push_back #define fi first #define se second #define debug(x) cerr<<#x << " := " << x << endl; #define bug cerr<<"-----------------------"<<endl; #define FOR(a, b, c) for(int a = b; a <= c; ++ a) typedef long long ll; typedef unsigned long long ull; typedef long double ld; typedef pair<int, int> pii; typedef pair<ll, ll> pll; const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9+7; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar(); while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar(); return x=f?-x:x; } /**********showtime************/ const int N = 2e5 + 5, M = 4e6 + 5;//M为节点个数,为Q*log(N) int root[N], lson[M], rson[M], value[M], tot = 0; ll sum[M]; const double eps = 1e-7; //建树 void build(int &x, int l, int r) { x = ++tot; value[x] = 0; sum[x] = 0; if(l == r) { return ; } int m = (l+r) >> 1; build(lson[x], l, m); build(rson[x], m+1, r); value[x] = value[lson[x]] + value[rson[x]]; } // 将某个历史版本p位置的值加v void update(int old, int &x, int p, int v, int l, int r) { x = ++tot; lson[x] = lson[old], rson[x] = rson[old], value[x] = value[old] + v, sum[x] = sum[old] + p; if(l == r) return ; int m = (l+r) >> 1; if(p <= m) update(lson[x], lson[x], p, v, l, m); else update(rson[x], rson[x], p, v, m+1, r); } //访问某个历史版本L到R的区间和 int query(int L, int R, int x, int l, int r) { if(L <= l && r <= R) return value[x]; int m = (l+r) >> 1, ans = 0; if(L <= m) ans += query(L, R, lson[x], l, m); if(R > m) ans += query(L, R, rson[x], m+1, r); return ans; } ll query2(int L, int R, int x, int l, int r) { if(L <= l && r <= R) return sum[x]; int m = (l+r) >> 1; ll ans = 0; if(L <= m) ans += query2(L, R, lson[x], l, m); if(R > m) ans += query2(L, R, rson[x], m+1, r); return ans; } const int maxn = 2e5+9; ll pre[maxn], a[maxn]; double cal(double val, int L, int R) { int hi = floor(val); int cnt = query(0, hi, root[R], 0, 100000) - query(0, hi, root[L-1], 0, 100000); double ss = (R - L + 1 - cnt) * val; ss += 1.0*query2(0, hi,root[R], 0, 100000) - query2(0, hi, root[L-1], 0, 100000); return ss; } int main(){ int n,m; scanf("%d%d", &n, &m); build(root[0], 0, 100000); for(int i=1; i<=n; i++) { scanf("%lld", &a[i]), pre[i] = pre[i-1] + a[i]; update(root[i-1], root[i], a[i], 1, 0, 100000); } while(m--) { int L, R, x, y; scanf("%d%d%d%d", &L, &R, &x, &y); ll ss = pre[R] - pre[L-1]; double nd = ss*1.0 / y *(y - x); double le = 0, ri = 1000000005, res = 0; //debug(nd); while(le + eps < ri) { double mid = (le + ri) / 2; if(cal(mid, L, R) <= nd) le = mid, res = mid; else ri = mid; } printf("%.10f\n", res); } return 0; }
J Symmetrical Painting
题意：
有n个矩形，宽度都为1，排列在坐标轴上，问消去一些矩形的一部分，使得原来的图形上下对称。这个对称图形最大可能面积。
思路：
有点类似扫描线的做法。
/* * @Author: chenkexing * @Date: 2019-08-16 15:34:14 * @Last Modified by: chenkexing * @Last Modified time: 2019-08-16 15:41:12 */ // #pragma GCC optimize(2) // #pragma GCC optimize(3) // #pragma GCC optimize(4) #include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> // #include<bits/extc++.h> // using namespace __gnu_pbds; using namespace std; #define pb push_back #define fi first #define se second #define debug(x) cerr<<#x << " := " << x << endl; #define bug cerr<<"-----------------------"<<endl; #define FOR(a, b, c) for(int a = b; a <= c; ++ a) typedef long long ll; typedef long double ld; typedef pair<int, int> pii; typedef pair<ll, ll> pll; const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; const int mod = 998244353; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar(); while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar(); return x=f?-x:x; } /**********showtime************/ const int maxn = 300009; pii a[maxn * 3]; int main(){ int n; scanf("%d", &n); int tot = 0; for(int i=1; i<=n; i++) { int le,ri; scanf("%d%d", &le, &ri); // 在底，中点，高上有可能取到极值。 a[++tot] = pii(2*le, 1); a[++tot] = pii(le + ri, -2); a[++tot] = pii(2*ri, 1); } sort(a+1, a+1+tot); ll ans = 0, sum = 0; ll cnt = a[1].se; for(int i=2; i<=tot; i++) { sum += cnt * (a[i].fi - a[i-1].fi); ans = max(ans, sum); cnt += a[i].se; } printf("%lld\n", ans); return 0; }