Sdoi2017试题泛做
Day1
[Sdoi2017]数字表格
推式子的莫比乌斯反演题。
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath> #define maxn 1000010
#define R register
const int mod = 1e9 + ;
int miu[maxn], fib[maxn], g[maxn], pr[maxn / ], prcnt, fp[maxn][];
bool vis[maxn];
inline int qpow(R int base, R int power)
{
R int ret = ;
for (; power; power >>= , base = 1ll * base * base % mod)
power & ? ret = 1ll * ret * base % mod : ;
return ret;
}
int main()
{
miu[] = ; fib[] = ; fib[] = g[] = g[] = ;
for (R int i = ; i < maxn; ++i)
{
fib[i] = (fib[i - ] + fib[i - ]) % mod; g[i] = ;
if (!vis[i]) pr[++prcnt] = i, miu[i] = -;
for (R int j = ; j <= prcnt && i * pr[j] < maxn; ++j)
{
vis[i * pr[j]] = ;
miu[i * pr[j]] = -miu[i];
if (i % pr[j] == )
{
miu[i * pr[j]] = ;
break;
}
}
}
for (R int i = ; i < maxn; ++i)
fp[i][] = qpow(fib[i], mod - ), fp[i][] = , fp[i][] = fib[i];
for (R int i = ; i < maxn; ++i)
for (R int j = i; j < maxn; j += i)
g[j] = 1ll * g[j] * fp[i][miu[j / i] + ] % mod;
for (R int i = ; i < maxn; ++i)
g[i] = (1ll * g[i - ] * g[i]) % mod;
R int T; scanf("%d", &T);
for (; T; --T)
{
R int n, m, ret = ; scanf("%d%d", &n, &m);
for (R int i = , j; i <= n && i <= m; i = j + )
{
j = std::min(n / (n / i), m / (m / i));
ret = 1ll * ret * qpow(1ll * g[j] * qpow(g[i - ], mod - ) % mod, 1ll * (n / i) * (m / i) % (mod - )) % mod;
}
printf("%d\n", ret);
}
return ;
}
D1T1
[Sdoi2017]树点涂色
LCT套线段树。
#include <cstdio>
#include <cstring> #define R register
#define maxn 100010
#define dmax(_a, _b) ((_a) > (_b) ? (_a) : (_b))
#define cmax(_a, _b) (_a < (_b) ? _a = (_b) : 0)
struct Edge {
Edge *next;
int to;
} *last[maxn], e[maxn << ], *ecnt = e;
inline void link(R int a, R int b)
{
*++ecnt = (Edge) {last[a], b}; last[a] = ecnt;
*++ecnt = (Edge) {last[b], a}; last[b] = ecnt;
}
int son[maxn], size[maxn], fa[maxn], top[maxn], dfn[maxn], pos[maxn], timer, dep[maxn], rig[maxn], n;
bool vis[maxn];
void dfs1(R int x)
{
vis[x] = ; size[x] = ; dep[x] = dep[fa[x]] + ;
for (R Edge *iter = last[x]; iter; iter = iter -> next)
if (!vis[iter -> to])
{
fa[iter -> to] = x;
dfs1(iter -> to);
size[x] += size[iter -> to];
size[son[x]] < size[iter -> to] ? son[x] = iter -> to : ;
}
}
void dfs2(R int x)
{
vis[x] = ; top[x] = son[fa[x]] == x ? top[fa[x]] : x; dfn[x] = ++timer; pos[timer] = x;
for (R Edge *iter = last[x]; iter; iter = iter -> next)
if (vis[iter -> to]) dfs2(iter -> to);
rig[x] = timer;
}
inline int getlca(R int a, R int b)
{
while (top[a] != top[b])
{
dep[top[a]] < dep[top[b]] ? b = fa[top[b]] : a = fa[top[a]];
}
return dep[a] < dep[b] ? a : b;
}
int tr[maxn << ], tag[maxn << ], ql, qr, qv;
inline void update(R int o) {tr[o] = dmax(tr[o << ], tr[o << | ]);}
inline void pushdown(R int o)
{
if (tag[o])
{
tr[o << ] += tag[o]; tag[o << ] += tag[o];
tr[o << | ] += tag[o]; tag[o << | ] += tag[o];
tag[o] = ;
}
}
void build(R int o, R int l, R int r)
{
if (l == r) { tr[o] = dep[pos[l]]; return ;}
R int mid = l + r >> ;
build(o << , l, mid); build(o << | , mid + , r);
update(o);
}
void modify(R int o, R int l, R int r)
{
if (ql <= l && r <= qr)
{
tag[o] += qv; tr[o] += qv; return ;
}
R int mid = l + r >> ;
pushdown(o);
if (ql <= mid) modify(o << , l, mid);
if (mid < qr) modify(o << | , mid + , r);
update(o);
}
int query(R int o, R int l, R int r)
{
if (ql <= l && r <= qr) return tr[o];
R int mid = l + r >> , ret = , tmp;
pushdown(o);
if (ql <= mid) tmp = query(o << , l, mid), cmax(ret, tmp);
if (mid < qr) tmp = query(o << | , mid + , r), cmax(ret, tmp);
update(o);
return ret;
}
struct Node *null;
struct Node {
Node *ch[], *fa, *mx;
inline bool type()
{
return fa -> ch[] == this;
}
inline bool check()
{
return fa -> ch[type()] == this;
}
inline void update()
{
mx = ch[] != null ? ch[] -> mx : this;
}
inline void rotate()
{
R Node *f = fa, *gf = f -> fa; R bool d = type();
(f -> ch[d] = ch[!d]) != null ? ch[!d] -> fa = f, : ;
(fa = gf), f -> check() ? gf -> ch[f -> type()] = this : ;
(ch[!d] = f) -> fa = this;
f -> update();
}
inline void splay()
{
for (; check(); rotate())
if (fa -> check())
{
(type() != fa -> type() ? this : fa) -> rotate();
}
update();
}
inline void access()
{
R Node *i = this, *j = null;
for (; i != null; i = (j = i) -> fa)
{
i -> splay();
// printf("i %d j %d\n", i - null, j - null);
if (i -> ch[] != null)
{
// printf("%d +1\n", i -> ch[1] -> mx - null);
ql = dfn[i -> ch[] -> mx - null]; qr = rig[i -> ch[] -> mx - null]; qv = ;
modify(, , n);
}
if (j != null)
{
// printf("%d -1\n", j - null);
ql = dfn[j -> mx - null]; qr = rig[j -> mx - null]; qv = -;
modify(, , n);
}
i -> ch[] = j;
}
}
} mem[maxn];
int main()
{
// freopen("in.in", "r", stdin);
// freopen("out.out", "w", stdout);
R int m; scanf("%d%d", &n, &m); null = mem;
for (R int i = ; i < n; ++i)
{
R int a, b; scanf("%d%d", &a, &b); link(a, b);
}
dfs1(); dfs2(); build(, , n);
for (R int i = ; i <= n; ++i) mem[i] = (Node) {{mem, mem}, mem + fa[i], mem + i};
for (; m; --m)
{
R int opt, x, y; scanf("%d%d", &opt, &x);
if (opt == ) (mem + x) -> access();
else if (opt == )
{
scanf("%d", &y);
R int lca = getlca(x, y), dx, dy, dlca;
ql = dfn[x]; qr = dfn[x];
dx = query(, , n);
ql = dfn[y]; qr = dfn[y];
dy = query(, , n);
ql = dfn[lca]; qr = dfn[lca];
dlca = query(, , n);
printf("%d\n", dx + dy - dlca * + );
}
else
{
ql = dfn[x]; qr = rig[x];
printf("%d\n", query(, , n));
}
}
return ;
}
D1T2
[Sdoi2017]序列计数
循环矩阵快速幂。
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <bitset>
#include <cstring> #define R register
#define maxn 20000010
const int mod = ;
int pr[maxn / ], prcnt;
std::bitset<maxn> vis;
typedef int Vector[];
Vector num1, num2, base;
int p;
void mul(R Vector A, R Vector B)
{
R Vector C; memset(C, , p << );
for (R int i = ; i < p; ++i) for (R int j = ; j < p; ++j)
C[(i + j) % p] = (C[(i + j) % p] + 1ll * A[i] * B[j]) % mod;
memcpy(A, C, p << );
}
int main()
{
R int n, m; scanf("%d%d%d", &n, &m, &p);
num1[] = ;
for (R int i = ; i <= m; ++i)
{
++num1[i % p];
if (!vis[i]) pr[++prcnt] = i, --num2[i % p];
for (R int j = ; j <= prcnt && i * pr[j] <= m; ++j)
{
vis[i * pr[j]] = ;
if (i % pr[j] == ) break;
}
}
for (R int i = ; i < p; ++i) num2[i] += num1[i];
memcpy(base, num1, p << );
memset(num1, , p << ); num1[] = ;
for (R int power = n; power; power >>= , mul(base, base))
power & ? mul(num1, base), : ;
memcpy(base, num2, p << );
memset(num2, , p << ); num2[] = ;
for (R int power = n; power; power >>= , mul(base, base))
power & ? mul(num2, base), : ;
printf("%d\n", (num1[] - num2[] + mod) % mod);
return ;
}
D1T3
Day2
[Sdoi2017]新生舞会
分数规划->二分+费用流。
#include <cstdio>
#include <cstring>
#include <algorithm> #define maxn 210
#define R register
#define cmin(_a, _b) (_a > (_b) ? _a = (_b) : 0)
typedef double db;
struct Edge {
Edge *next, *rev;
int from, to, cap;
db cost;
} *last[maxn], *prev[maxn], e[maxn * maxn * ], *ecnt = e;
inline void link(R int a, R int b, R int w, R db c)
{
*++ecnt = (Edge) {last[a], ecnt + , a, b, w, c}; last[a] = ecnt;
*++ecnt = (Edge) {last[b], ecnt - , b, a, , -c}; last[b] = ecnt;
}
const db inf = 1e9;
int a[maxn][maxn], b[maxn][maxn], n, q[maxn * ], s, t;
db ans, dis[maxn];
bool inq[maxn];
inline bool spfa()
{
for (R int i = ; i <= t; ++i) dis[i] = inf;
R int head = , tail = ; q[] = s;
while (head < tail)
{
R int now = q[++head]; inq[now] = ;
for (R Edge *iter = last[now]; iter; iter = iter -> next)
if (iter -> cap && dis[iter -> to] > dis[now] + iter -> cost)
{
dis[iter -> to] = dis[now] + iter -> cost;
prev[iter -> to] = iter;
!inq[iter -> to] ? inq[q[++tail] = iter -> to] = : ;
}
}
return dis[t] != inf;
}
inline void mcmf()
{
R int x = 0x7fffffff;
for (R Edge *iter = prev[t]; iter; iter = prev[iter -> from]) cmin(x, iter -> cap);
for (R Edge *iter = prev[t]; iter; iter = prev[iter -> from])
{
iter -> cap -= x;
iter -> rev -> cap += x;
ans += iter -> cost * x;
}
}
inline db val(R db k)
{
memset(last, , (t + ) << ); ecnt = e; ans = ;
for (R int i = ; i <= n; ++i)
{
link(s, i, , ); link(i + n, t, , );
for (R int j = ; j <= n; ++j)
link(i, j + n, , k * b[i][j] - a[i][j]);
}
while (spfa()) mcmf();
return -ans;
}
int main()
{
scanf("%d", &n);
for (R int i = ; i <= n; ++i) for (R int j = ; j <= n; ++j) scanf("%d", &a[i][j]);
for (R int i = ; i <= n; ++i) for (R int j = ; j <= n; ++j) scanf("%d", &b[i][j]);
s = ; t = n << | ;
R db left = , right = 1e4;
while (right - left > 1e-)
{
R db mid = (left + right) * 0.5;
if (val(mid) > ) left = mid;
else right = mid;
}
printf("%.6lf\n", left);
return ;
}
D2T1
[Sdoi2017]硬币游戏
很神的题。一开始只会6方的高斯消元。后来看了题解，计f[i]为i获胜的概率。f[0]表示没人获胜的概率。若我们计S_i表示i获胜的字符串的集合，S_0表示没有人获胜的字符串的集合。显然，这些集合的大小都是正无穷的。但我们只需要知道它们之间的比率。
我们以样例为例。
3 3
THT
HTT
TTH
首先，我们可以确定的是在所有的S_0的后面加上`THT`肯定游戏就结束了。
但问题是赢的不一定是第一个人。
比如，若S_0里最后两位如果是`HT`，那么此时就会在加上第一个`T`时就结束游戏了。同理，最后一位如果是`T`的话，那么加上`TH`以后就结束游戏了。
那么我们可以考虑，在所有的S_0集合后加上`THT`这个后缀，得到的字符串集合可以表示为：{1获胜的集合，2获胜的集合+`HT`，3获胜的集合+`T`}。
（注意：这里的`HT`和`T`均为`THT`的某个后缀）
所以可以得到这样的式子：0.125 * f_0 = f_1 + f_2 * 0.25 + f_3 * 0.5。然后对于每个串都能得到一个这样的式子，再加上所有人获胜的概率和为1，我们就得到了一个n+1个未知数，有n+1个方程的方程组。用高斯消元解决即可。
如何得到上面的系数？我们发现是会有上面的那种情况只有可能是i的某个前缀和j的某个后缀相等。所以可以用KMP把两个串拼起来来找到每个前缀和后缀相等的长度。
#include <cstdio>
#include <cmath>
#include <algorithm> #define R register
#define maxn 310
typedef double db;
char s[maxn][maxn], st[maxn << ];
db a[maxn][maxn], x[maxn], pw[maxn];
int fa[maxn << ], m;
inline db calc(R int a, R int b)
{
for (R int i = ; i <= m; ++i) st[i] = s[a][i]; st[m + ] = '';
for (R int i = ; i <= m; ++i) st[i + m + ] = s[b][i];
fa[] = ;
R int p = ;
// puts(st + 1);
for (R int i = ; i <= m + m + ; ++i)
{
while (p && st[p + ] != st[i]) p = fa[p];
st[p + ] == st[i] ? ++p : ;
fa[i] = p;
// printf("%d\n", fa[i]);
}
R db ans = ;
// printf("%d %d\n", a, b);
while (p)
{
// printf("p = %d\n", p);
ans += pw[m - p];
p = fa[p];
}
// printf("%.2lf\n", ans);
return ans;
}
int main()
{
R int n; scanf("%d%d", &n, &m);
for (R int i = ; i <= n; ++i) scanf("%s", s[i] + );
pw[] = ; for (R int i = ; i <= m; ++i) pw[i] = pw[i - ] * 0.5;
for (R int i = ; i <= n; ++i)
{
a[i][] = -pw[m];
a[][i] = ;
for (R int j = ; j <= n; ++j)
{
// if (i == j) continue;
a[i][j] += calc(i, j);
}
}
a[][n + ] = ; for (R int i = ; i <= n; ++i)
{
if (fabs(a[i][i]) < 1e-)
{
for (R int j = i + ; j <= n; ++j)
if (fabs(a[j][i]) > fabs(a[i][i]))
{
for (R int k = i; k <= n + ; ++k) std::swap(a[i][k], a[j][k]);
break;
}
}
for (R int j = i + ; j <= n; ++j)
{
R db temp = a[j][i] / a[i][i];
for (R int k = i; k <= n + ; ++k)
a[j][k] -= a[i][k] * temp;
}
}
x[n] = a[n][n + ] / a[n][n];
// fprintf(stderr, "%.2lf %.2lf\n", a[n][n], a[n][n + 1]);
for (R int i = n - ; i; --i)
{
R db tmp = a[i][n + ];
for (R int j = i + ; j <= n; ++j)
tmp -= a[i][j] * x[j];
x[i] = tmp / a[i][i];
}
for (R int i = ; i <= n; ++i) printf("%.10lf\n", x[i]);
return ;
}
D2T2
[Sdoi2017]相关分析
把式子强拆开来发现维护区间x_i的和，y_i的和，x_i^2的和，x_i*y_i的和。还需要资磁x,y分别区间加，区间覆盖成一个等差数列（因为公差为1所以比较容易维护）。
这些线段树都可以做，细节比较多吧，调的时候最好还是靠对拍。
#include <cstdio> #define R register
#define maxn 1048576
int xi[maxn], yi[maxn];
typedef long long ll;
typedef double db;
struct data {
db x, y, xy, xx;
inline data operator + (const data &that) const {return (data) {x + that.x, y + that.y, xy + that.xy, xx + that.xx};}
inline data operator += (const data &that) {x += that.x; y += that.y; xy += that.xy; xx += that.xx; }
};
struct Seg {
int tag_set_x, tag_set_y, tag_add_x, tag_add_y;
data x;
} tr[maxn];
ll i2[maxn];
inline void update(R int o)
{
tr[o].x = tr[o << ].x + tr[o << | ].x;
}
void data_add(R data &x, R int s, R int t, R int len)
{
x.xx += * s * x.x + (db) s * s * len;
x.xy += s * x.y + t * x.x + (db) s * t * len;
x.x += (db) s * len;
x.y += (db) t * len;
}
void data_set(R data &x, R int s, R int t, R int l, R int r)
{
R ll si = (db) (l + r) * (r - l + ) / ;
x.x = (db) s * (r - l + ) + si;
x.y = (db) t * (r - l + ) + si;
x.xx = (db) s * s * (r - l + ) + * s * si + i2[r] - i2[l - ];
x.xy = i2[r] - i2[l - ] + (db) s * t * (r - l + ) + si * (s + t);
// printf("%lld %lld %lld %lld\n", x.x, x.y, x.xx, x.xy);
}
void pushdown(R int o, R int l, R int r)
{
R int mid = l + r >> ;
if (tr[o].tag_set_x != - || tr[o].tag_set_y != -)
{
data_set(tr[o << ].x, tr[o].tag_set_x, tr[o].tag_set_y, l, mid);
data_set(tr[o << | ].x, tr[o].tag_set_x, tr[o].tag_set_y, mid + , r); tr[o << ].tag_set_x = tr[o].tag_set_x;
tr[o << ].tag_set_y = tr[o].tag_set_y;
tr[o << | ].tag_set_x = tr[o].tag_set_x;
tr[o << | ].tag_set_y = tr[o].tag_set_y;
tr[o << ].tag_add_x = ;
tr[o << | ].tag_add_x = ;
tr[o << ].tag_add_y = ;
tr[o << | ].tag_add_y = ; tr[o].tag_set_x = tr[o].tag_set_y = -;
}
if (tr[o].tag_add_x || tr[o].tag_add_y)
{
data_add(tr[o << ].x, tr[o].tag_add_x, tr[o].tag_add_y, mid - l + );
data_add(tr[o << | ].x, tr[o].tag_add_x, tr[o].tag_add_y, r - mid); tr[o << ].tag_add_x += tr[o].tag_add_x;
tr[o << ].tag_add_y += tr[o].tag_add_y;
tr[o << | ].tag_add_x += tr[o].tag_add_x;
tr[o << | ].tag_add_y += tr[o].tag_add_y; tr[o].tag_add_x = tr[o].tag_add_y = ;
}
}
void build(R int o, R int l, R int r)
{
tr[o].tag_set_x = tr[o].tag_set_y = -; tr[o].tag_add_x = tr[o].tag_add_y = ;
if (l == r)
{
tr[o].x = (data) {xi[l], yi[l], (db) xi[l] * yi[l], (db) xi[l] * xi[l]};
return ;
}
R int mid = l + r >> ;
build(o << , l, mid); build(o << | , mid + , r);
update(o);
}
data ret; int ql, qr, s, t;
void query(R int o, R int l, R int r)
{
if (ql <= l && r <= qr)
{
ret += tr[o].x; return ;
}
R int mid = l + r >> ;
pushdown(o, l, r);
if (ql <= mid) query(o << , l, mid);
if (mid < qr) query(o << | , mid + , r);
update(o);
}
void modify_add(R int o, R int l, R int r)
{
if (ql <= l && r <= qr)
{
tr[o].tag_add_x += s;
tr[o].tag_add_y += t;
data_add(tr[o].x, s, t, r - l + );
return ;
}
R int mid = l + r >> ;
pushdown(o, l, r);
if (ql <= mid) modify_add(o << , l, mid);
if (mid < qr) modify_add(o << | , mid + , r);
update(o);
}
void modify_set(R int o, R int l, R int r)
{
if (ql <= l && r <= qr)
{
tr[o].tag_set_x = s;
tr[o].tag_set_y = t;
tr[o].tag_add_x = tr[o].tag_add_y = ;
data_set(tr[o].x, s, t, l, r);
return ;
}
R int mid = l + r >> ;
pushdown(o, l, r);
if (ql <= mid) modify_set(o << , l, mid);
if (mid < qr) modify_set(o << | , mid + , r);
update(o);
}
int main()
{
R int n, m; scanf("%d%d", &n, &m);
for (R int i = ; i <= n; ++i) scanf("%d", &xi[i]);
for (R int i = ; i <= n; ++i) scanf("%d", &yi[i]);
for (R int i = ; i <= n; ++i) i2[i] = i2[i - ] + (db) i * i;
build(, , n);
for (; m; --m)
{
R int opt, l, r; scanf("%d%d%d", &opt, &ql, &qr);
if (opt == )
{
ret = (data) {, , , };
query(, , n); l = ql; r = qr;
R db _x = (db) ret.x / (r - l + ), _y = (db) ret.y / (r - l + );
printf("%.10lf\n", (ret.xy - _y * ret.x - _x * ret.y + _x * _y * (r - l + )) / (ret.xx - * _x * ret.x + _x * _x * (r - l + )));
}
else if (opt == )
{
scanf("%d%d", &s, &t);
modify_add(, , n);
}
else
{
scanf("%d%d", &s, &t);
modify_set(, , n);
}
}
return ;
}
D2T3
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