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垫底进队之后第一次集训。
Maximum Multiple
3 的倍数拆 1,1,1 ，4 的倍数拆 1,1,2 ，优先拆 3 的倍数。
#include <stdio.h>
long long t, n;
int main()
{
for (scanf("%lld", &t); t--;
printf("%lld\n", n % 3 == 0 ? n * n * n / 27 : n % 4 == 0 ? n * n * n / 32 : -1))
scanf("%lld", &n);
}
Balanced Sequence
具体思想是把每个串按删去可以自匹配的括号后的剩余串（形如 )))(( ）排序：
左括号多于右括号的串排在左括号少于右括号的串前面；
同类型的串中，左括号多于右括号的串右括号少的放前面，右括号多于左括号的串左括号少的放前面。
现场谜之排序 WA 了十一发…
#include <bits/stdc++.h>
using namespace std;
const int N = 100009;
struct Node
{
int l, r;
bool operator<(const Node &t) const
{
return r < l ? (t.r < t.l ? r < t.r : 1) : (t.r < t.l ? 0 : l > t.l);
}
} v[N];
char s[N];
int t, n, ans;
int main()
{
for (scanf("%d", &t); t--;)
{
scanf("%d", &n);
for (int i = ans = 0; i < n; ++i)
{
scanf("%s", s);
for (int j = v[i].l = v[i].r = 0; s[j]; ++j)
{
if (s[j] == '(')
++v[i].l;
else if (v[i].l)
--v[i].l, ++ans;
else
++v[i].r;
}
}
sort(v, v + n);
for (int i = 0, cnt = 0; i < n; ++i)
{
ans += min(cnt, v[i].r);
cnt += v[i].l - min(cnt, v[i].r);
}
printf("%d\n", ans * 2);
}
}
Triangle Partition
把所有点按照横坐标-纵坐标的字典序排序，选取依次相邻三项。
#include <bits/stdc++.h>
using namespace std;
struct Coord : pair<int, int>
{
int id;
} p[10009];
int t, n;
int main()
{
for (scanf("%d", &t); t--;)
{
scanf("%d", &n);
for (int i = 0; i < 3 * n; ++i)
scanf("%d%d", &p[i].first, &p[i].second), p[i].id = i;
sort(p, p + 3 * n);
for (int i = 0; i < 3 * n; ++i)
printf("%d%c", p[i].id + 1, i % 3 != 2 ? ' ' : '\n');
}
}
Distinct Values
队友写的。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn = 1e5 + 9;
pair<int, int> a[maxn];
int ans[maxn];
int t, n, m;
priority_queue<int, vector<int>, greater<int>> q;
int main()
{
scanf("%d", &t);
while (t--)
{
scanf("%d%d", &n, &m);
memset(ans, 0, sizeof(ans));
if (m == 0)
{
for (int i = 1; i < n; i++)
printf("1 ");
printf("1\n");
continue;
}
for (int i = 1; i <= m; i++)
scanf("%d%d", &a[i].first, &a[i].second);
sort(a + 1, a + m + 1);
while (!q.empty())
q.pop();
// for (int i=1;i<=m;i++) printf("!%d %d\n",a[i].first,a[i].second);
for (int i = 1; i < a[1].first; i++)
ans[i] = 1;
for (int i = a[1].first; i <= a[1].second; i++)
ans[i] = i - a[1].first + 1;
int l = a[1].first, r = a[1].second, mx = ans[a[1].second];
for (int i = 2; i <= m; i++)
{
if (a[i].second <= r)
continue;
if (a[i].first <= r)
{
for (int j = l; j < a[i].first; j++)
q.push(ans[j]);
for (int j = r + 1; j <= a[i].second; j++)
if (!q.empty())
ans[j] = q.top(), q.pop();
else
ans[j] = ++mx;
}
else
{
while (!q.empty())
q.pop();
mx = 0;
for (int j = a[i].first; j <= a[i].second; j++)
ans[j] = ++mx;
}
l = a[i].first;
r = a[i].second;
}
for (int i = 1; i <= n; i++)
if (ans[i] == 0)
ans[i] = 1;
for (int i = 1; i < n; i++)
printf("%d ", ans[i]);
printf("%d\n", ans[n]);
}
}
Maximum Weighted Matching
//假装会写
Period Sequence
//假装会写
Chiaki Sequence Revisited
队友$O(\log\log N)$的二分强行跑掉！太强啦！
#include <cstdio>
typedef long long ll;
const ll mod = 1e9 + 7;
int t;
ll n, v, ans, u;
inline ll geti(ll x)
{
ll l = x / 2 - 1, r = x;
while (r - l > 1)
{
ll mid = (l + r) / 2, k = mid;
ll g = 0;
while (k > 0)
{
g += k;
k /= 2;
}
// printf("%lld %lld %lld %lld\n",l,r,mid,g);
if (g < x)
l = mid;
else
r = mid;
}
return r;
}
inline ll cal(ll x)
{
ll ret = 0, t = 1;
u = 0;
while (x > 0)
{
if (x & 1)
ret = (ret + ((x % mod) * ((x + 1) / 2 % mod) % mod) * t) % mod;
else
ret = (ret + (((x + 1) % mod) * ((x / 2) % mod) % mod) * t) % mod;
//printf("ret=%lld\n",ret);
u += x;
t = (t * 2) % mod;
x /= 2;
}
return ret;
}
int main()
{
scanf("%d", &t);
while (t--)
{
scanf("%lld", &n);
v = geti(n - 1);
// printf("%lld ",v);
ans = cal(v - 1);
ans = (ans + v * (n - 1 - u) + 1) % mod;
printf("%lld\n", ans);
}
}
RMQ Similar Sequence
学习一个一队大爷的分治做法。
找区间 max 递归计算即可，这里是用单调栈建了一棵笛卡尔树。
加上快速读入，快的一批。
#include <bits/stdc++.h>
#define cin kin
using namespace std;
typedef int ll;
const int N = 1e6 + 7;
struct Istream
{
char b[20 << 20], *i, *e;
Istream(FILE *in) : i(b), e(b + fread(b, sizeof(*b), sizeof(b), in)) {}
Istream &operator>>(ll &val)
{
while (*i < '0')
++i;
for (val = 0; *i >= '0'; ++i)
val = (val << 3) + (val << 1) + *i - '0';
return *this;
}
} kin(stdin);
struct Mod
{
const ll M;
Mod(ll M) : M(M) {}
ll mul(long long a, ll b) const { return a * b % M; }
};
struct Factorial : Mod
{
vector<ll> fac, ifac;
Factorial(int N, ll M) : fac(N, 1), ifac(N, 1), Mod(M)
{
for (int i = 2; i < N; ++i)
fac[i] = mul(fac[i - 1], i), ifac[i] = mul(M - M / i, ifac[M % i]);
for (int i = 2; i < N; ++i)
ifac[i] = mul(ifac[i], ifac[i - 1]);
}
ll c(int n, int m) { return mul(mul(fac[n], ifac[m]), ifac[n - m]); }
} f(N, 1e9 + 7);
struct Node
{
int val, lc, rc;
} tr[N];
ll t, n, s;
void dfs(int l, int r, int m)
{
if (l > r)
return;
if (l < m && m < r)
s = f.mul(s, f.c(r - l, m - l));
dfs(l, m - 1, tr[m].lc), dfs(m + 1, r, tr[m].rc);
}
int main()
{
for (cin >> t; t--; cout << s << '\n')
{
cin >> n;
vector<int> stak;
for (int i = 0; i < n; ++i)
{
for (cin >> tr[i].val; !stak.empty() && tr[stak.back()].val < tr[i].val; stak.pop_back())
tr[i].lc = stak.back();
if (!stak.empty())
tr[stak.back()].rc = i;
stak.push_back(i);
}
s = f.mul(f.mul(n, f.ifac[2]), f.ifac[n]), dfs(0, n - 1, stak.front());
}
}
Lyndon Substring
//假装会写
Turn Off The Light
//假装会写
Time Zone
读到的 double 乘十后要四舍五入，且 istringstream 速度太慢过不了。
#include <stdio.h>
char s[9];
double d;
int t, a, b;
int main()
{
for (scanf("%d", &t); t--; printf("%02d:%02d\n", (a + b / 60) % 24, b % 60))
{
scanf("%d%d%s", &a, &b, &s);
sscanf(s + 3, "%lf", &d);
if (d -= 8, d < 0)
d += 24;
b += 6 * (int)(d * 10 + 0.5);
}
}