2022NOIPA层联测32
有人知道 for(int x : vec) ...这东西noip考场上能不能用啊？今天上午想用来着结果过不了编译？？
B. 鸟之诗 (air)
赛时只会全排列y的质因数，并且认为用a和b把x和y乘起来麻烦死了……丝毫都没有往正解那方面想……
5 pts
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 3;
const int mod = 998244353;
int p[10], n, m, fz[10], num, b[10];
ll x, y, ans;
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
inline int gcd(int a, int b)
{
while(b^=a^=b^=a%=b);
return a;
}
inline int lcm(int x, int y)
{
int d = gcd(x, y);
return x / d * y;
}
void calc()
{
int d = b[1];
for(int i=2; i<=n; i++)
{
d = gcd(d, b[i]);
}
if(d != x) return;
d = b[1];
for(int i=2; i<=n; i++)
{
d = lcm(d, b[i]);
}
if(d != y) return;
ans++;
}
void dfs(int a)
{
if(a > n)
{
calc(); return;
}
for(int i=1; i<=num; i++)
{
if(fz[i] < x) continue;
b[a] = fz[i];
dfs(a+1);
}
}
int main()
{
freopen("air.in", "r", stdin);
freopen("air.out", "w", stdout);
m = read(); n = read();
p[1] = 2, p[2] = 3, p[3] = 5, p[4] = 7, p[5] = 11, p[6] = 13;
x = 1, y = 1;
for(int i=1; i<=m; i++)
{
int l = read();
int res = 1;
for(int j=1; j<=l; j++) res *= p[i];
x = x * res;
}
for(int i=1; i<=m; i++)
{
int l = read();
int res = 1;
for(int j=1; j<=l; j++) res *= p[i];
y = y * res;
}
for(int i=1; i<=sqrt(y); i++)
{
if(y % i == 0)
{
fz[++num] = i;
if(i * i != y) fz[++num] = y / i;
}
}
dfs(1);
ans %= mod;
printf("%lld\n", ans);
return 0;
}code
#include <bits/stdc++.h>
using namespace std;
//#define int long long
typedef long long ll;
const int maxn = 1e6 + 7;
const int mod = 998244353;
ll m, n, a[maxn], b[maxn];
ll ans = 1;
inline ll read()
{
ll x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
ll qpow(ll a, ll b)
{
ll ans = 1;
while(b)
{
if(b & 1) ans = ans * a % mod;
a = a * a % mod;
b >>= 1;
}
return ans;
}
int main()
{
freopen("air.in", "r", stdin);
freopen("air.out", "w", stdout);
m = read(); n = read(); n %= (mod-1);
for(int i=1; i<=m; i++) a[i] = read();
for(int i=1; i<=m; i++) b[i] = read();
for(int i=1; i<=m; i++)
{
if(a[i] == b[i]) continue;
ll res = (qpow(b[i]-a[i]+1, n)-2ll*qpow(b[i]-a[i], n)%mod+qpow(b[i]-a[i]-1, n))%mod;
res = (res + mod) % mod;
ans = ans * res % mod;
}
printf("%lld\n", ans);
return 0;
}
D. 无双挑战 (challenge)
把树上距离拆开，不好算的就只有dep[lca]求和，解决方法是从upd这个点到根区间加1，query统计点到根区间求和(把加的1合起来就是深度嘛)，严格大于就把相等的拆出来一起算。
这题本来没有坑，So我给自己挖了坑跳了进去：区间修改的线段树不打lazy，在范围内的时候t[x].sum+=v没乘区间长度，把 j 写成 i ，(l+r)>>1写成(l+1)>>1，结果就是就这题还调了一下午！？并且因为long long没开全先炸没了菊花图的部分分，正解又挂了一次96……
我这才多久没写线段树区间求和……
把上次联测的sily错误也在这里说一下好了(那天博客园登不上去)：那个最短路的题就算不开根号也都知道是三分，我写了个二分就&*@#￥，我不知道为什么当我看到x + d/(x+1)这个函数时居然以为它单调递增？？用脚想都说不通，它单点递增我还分个什么鬼！？
我这才多久没做数学T……
顺便说一下今天的T3他给的式子直接用就30结果我从1循环到了m似乎根本就不知道区间划分是从二维开始的，过不了样例以为是他的提示不能推广……
32 pts
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 4;
const int mod = 998244353;
const int inf = 0x3f3f3f3f;
int n, m, dep[maxn], top[maxn], siz[maxn], fa[maxn], son[maxn];
int v[maxn];
ll ans;
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
struct node
{
int nxt, to;
}e[maxn<<1];
int head[maxn], len;
void add(int x, int y)
{
e[++len].to = y; e[len].nxt = head[x];
head[x] = len;
}
void find_heavy_edge(int u, int fat, int depth)
{
fa[u] = fat;
dep[u] = depth;
siz[u] = 1;
son[u] = 0;
int maxsize = 0;
for(int i=head[u]; i; i=e[i].nxt)
{
int v = e[i].to;
if(v == fat) continue;
find_heavy_edge(v, u, depth+1);
siz[u] += siz[v];
if(siz[v] > maxsize)
{
maxsize = siz[v];
son[u] = v;
}
}
}
void connect_heavy_edge(int u, int ancestor)
{
top[u] = ancestor;
if(son[u])
{
connect_heavy_edge(son[u], ancestor);
}
for(int i=head[u]; i; i=e[i].nxt)
{
int v = e[i].to;
if(v == fa[u] || v == son[u]) continue;
connect_heavy_edge(v, v);
}
}
int LCA(int x, int y)
{
while(top[x] != top[y])
{
if(dep[top[x]] < dep[top[y]]) swap(x, y);
x = fa[top[x]];
}
if(dep[x] > dep[y]) swap(x, y);
return x;
}
ll qpow(ll a, ll b)
{
ll ans = 1;
while(b)
{
if(b & 1) ans = ans * a % mod;
a = a * a % mod;
b >>= 1;
}
return ans;
}
ll sum[maxn];
struct node2
{
int id, v;
bool operator < (const node2 &T) const
{
return v < T.v;
}
}p[maxn];
int find(int mv)
{
int l = 1, r = n + 1;
while(l < r)
{
int mid = (l + r) >> 1;
if(p[mid].v > mv) r = mid;
else l = mid + 1;
}
return l;
}
void solveflwr()
{
ll res = 0;
for(int i=2; i<=n; i++)
{
if(v[i] > v[1]) ans++;
else if(v[1] > v[i]) res = (res + m - v[i]) % mod;
}
ans = ans * (m - v[1]) % mod;
for(int i=1; i<n; i++) p[i].v = v[i+1];
sort(p+1, p+n); p[n].v = inf;
for(int i=1; i<n; i++)
{
int pos = find(p[i].v);
ans = (ans+1ll*(m-p[i].v)*(n-pos+n-pos)%mod)%mod;
}
ans = (ans + res) % mod;
ll t = qpow(1ll*n*(n-1)%mod, mod-2);
ans = ans * t % mod;
printf("%lld\n", ans);
exit(0);
}
int main()
{
freopen("challenge.in", "r", stdin);
freopen("challenge.ans", "w", stdout);
n = read(); m = read(); bool fl = 1;
for(int i=1; i<=n; i++) v[i] = read();
for(int i=1; i<n; i++)
{
int u = read(), v = read();
if(u > v) swap(u, v);
if(u != 1) fl = 0;
add(u, v); add(v, u);
}
if(fl) solveflwr();
find_heavy_edge(1, 0, 1);
connect_heavy_edge(1, 1);
for(int i=1; i<=n; i++)
{
for(int j=1; j<=n; j++)
{
if(v[j] <= v[i]) continue;
int lca = LCA(i, j);
int dis = dep[i] - dep[lca] + dep[j] - dep[lca];
ans = (ans + (ll)(m - v[i]) * dis % mod) % mod;
}
}
ll t = qpow(1ll*n*(n-1)%mod, mod-2);
ans = ans * t % mod;
printf("%lld\n", ans);
return 0;
}code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
//#define int ll
const int maxn = 8e5 + 4;
const int mod = 998244353;
const int inf = 0x3f3f3f3f;
int n, m, sumdep, ans, flly;
int dep[maxn], siz[maxn], son[maxn], fa[maxn], top[maxn], dfn[maxn], dfn_c;
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
struct node
{
int a, id;
bool operator < (const node &T) const
{
return a > T.a;
}
}p[maxn];
struct node2
{
int nxt, to;
}e[maxn<<1];
int head[maxn], len;
void add_e(int x, int y)
{
e[++len].to = y; e[len].nxt = head[x];
head[x] = len;
}
inline void add(int &x, int y) {x=(x+y)%mod;}
struct seg
{
struct tree
{
int sum, lazy;
}t[maxn<<2];
void pushup(int x)
{
t[x].sum = (t[x<<1].sum + t[x<<1|1].sum) % mod;
}
void pushdown(int x, int l, int r)
{
int ls = x<<1, rs = x<<1|1, mid = (l + r) >> 1;
add(t[ls].lazy, t[x].lazy);
add(t[rs].lazy, t[x].lazy);
add(t[ls].sum, 1ll*(mid-l+1)*t[x].lazy%mod);
add(t[rs].sum, 1ll*(r-mid)*t[x].lazy%mod);
t[x].lazy = 0;
}
void update(int x, int l, int r, int L, int R, int v)
{
if(L <= l && r <= R)
{
add(t[x].sum, 1ll*v*(r-l+1)%mod); add(t[x].lazy, v);
return;
}
if(t[x].lazy) pushdown(x, l, r);
int mid = (l + r) >> 1;
if(L <= mid) update(x<<1, l, mid, L, R, v);
if(R > mid) update(x<<1|1, mid+1, r, L, R, v);
pushup(x);
}
int query(int x, int l, int r, int L, int R)
{
if(L <= l && r <= R) return t[x].sum;
if(t[x].lazy) pushdown(x, l, r);
int mid = (l + r) >> 1, ans = 0;
if(L <= mid) add(ans, query(x<<1, l, mid, L, R));
if(R > mid) add(ans, query(x<<1|1, mid+1, r, L, R));
return ans;
}
}t;
void find_heavy_edge(int u, int fat, int depth)
{
fa[u] = fat;
dep[u] = depth;
siz[u] = 1;
son[u] = 0;
int maxsize = 0;
for(int i=head[u]; i; i=e[i].nxt)
{
int v = e[i].to;
if(v == fat) continue;
find_heavy_edge(v, u, depth+1);
siz[u] += siz[v];
if(siz[v] > maxsize)
{
maxsize = siz[v];
son[u] = v;
}
}
}
void connect_heavy_edge(int u, int ancestor)
{
top[u] = ancestor;
dfn[u] = ++dfn_c;
if(son[u])
{
connect_heavy_edge(son[u], ancestor);
}
for(int i=head[u]; i; i=e[i].nxt)
{
int v = e[i].to;
if(v == fa[u] || v == son[u]) continue;
connect_heavy_edge(v, v);
}
}
void upd_ans(int x)
{
while(top[x] != top[1])
{
t.update(1, 1, n, dfn[top[x]], dfn[x], 1);
x = fa[top[x]];
}
t.update(1, 1, n, dfn[1], dfn[x], 1);
}
int get_ans(int x)
{
int ans = 0;
while(top[x] != top[1])
{
add(ans, t.query(1, 1, n, dfn[top[x]], dfn[x]));
x = fa[top[x]];
}
add(ans, t.query(1, 1, n, dfn[1], dfn[x]));
return ans;
}
ll qpow(ll a, ll b)
{
ll ans = 1;
while(b)
{
if(b & 1) ans = ans * a % mod;
a = a * a % mod;
b >>= 1;
}
return ans;
}
signed main()
{
freopen("challenge.in", "r", stdin);
freopen("challenge.out", "w", stdout);
n = read(), m = read();
for(int i=1; i<=n; i++)
{
p[i].a = read(), p[i].id = i;
}
for(int i=1; i<n; i++)
{
int u = read(), v = read();
add_e(u, v); add_e(v, u);
}
sort(p+1, p+1+n);
find_heavy_edge(1, 0, 1);
connect_heavy_edge(1, 1);
for(int i=1; i<=n; )
{
int l = i, r = i;
while(r < n && p[r+1].a == p[l].a) r++;
for(int j=l; j<=r; j++)
{
ans = 0;
add(ans, 1ll*dep[p[j].id]*(l-1)%mod);
add(ans, sumdep);
add(ans, (mod-2ll*get_ans(p[j].id)%mod)%mod);
add(flly, 1ll*ans*(m-p[j].a)%mod);
}
for(int j=l; j<=r; j++) add(sumdep, dep[p[j].id]);
for(int j=l; j<=r; j++) upd_ans(p[j].id);
i = r + 1;
}
flly = 1ll * flly * qpow(1ll*n*(n-1)%mod, mod-2) % mod;
printf("%d\n", flly);
return 0;
}
[LNOI2014]LCA
双倍经验？用到了差分，序列上差分和树上差分形式好像哦，然而我没有想到差分直接写了一个莫队然后T飞了555……还有就是记得多取几个mod，尤其是用到减法的时候(我又出负数了*)
code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 5e4 + 3;
const int mod = 201314;
int n, m, ans[maxn], sz;
int dep[maxn], fa[maxn], siz[maxn], son[maxn], dfn[maxn], dfn_c, top[maxn];
inline void add(int &x, int y) {x=(x+y)%mod;}
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')
{
f = -1;
}
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch^48);
ch = getchar();
}
return x * f;
}
struct que
{
int id, opt, z;
}q[maxn];
vector<que> v[maxn];
struct seg
{
struct tree
{
int sum, lazy;
}t[maxn<<2];
void pushup(int x)
{
t[x].sum = (t[x<<1].sum + t[x<<1|1].sum) % mod;
}
void pushdown(int x, int l, int r)
{
int ls = x<<1, rs = x<<1|1, mid = (l + r) >> 1;
add(t[ls].lazy, t[x].lazy);
add(t[rs].lazy, t[x].lazy);
add(t[ls].sum, 1ll*(mid-l+1)*t[x].lazy%mod);
add(t[rs].sum, (r-mid)*t[x].lazy%mod);
t[x].lazy = 0;
}
void update(int x, int l, int r, int L, int R, int v)
{
if(L <= l && r <= R)
{
add(t[x].sum, (r-l+1)*v%mod); add(t[x].lazy, v);
return;
}
if(t[x].lazy) pushdown(x, l, r);
int mid = (l + r) >> 1;
if(L <= mid) update(x<<1, l, mid, L, R, v);
if(R > mid) update(x<<1|1, mid+1, r, L, R, v);
pushup(x);
}
int query(int x, int l, int r, int L, int R)
{
if(L <= l && r <= R) return t[x].sum;
if(t[x].lazy) pushdown(x, l, r);
int mid = (l + r) >> 1, ans = 0;
if(L <= mid) add(ans, query(x<<1, l, mid, L, R));
if(R > mid) add(ans, query(x<<1|1, mid+1, r, L, R));
return ans;
}
}t;
struct node
{
int nxt, to;
}e[maxn];
int head[maxn], len;
void add_e(int x, int y)
{
e[++len].to = y; e[len].nxt = head[x];
head[x] = len;
}
void find_heavy_edge(int u, int fat, int depth)
{
fa[u] = fat;
dep[u] = depth;
siz[u] = 1;
son[u] = 0;
int maxsize = 0;
for(int i=head[u]; i; i=e[i].nxt)
{
int v = e[i].to;
find_heavy_edge(v, u, depth+1);
siz[u] += siz[v];
if(siz[v] > maxsize)
{
maxsize = siz[v];
son[u] = v;
}
}
}
void connect_heavy_edge(int u, int ancestor)
{
top[u] = ancestor;
dfn[u] = ++dfn_c;
if(son[u])
{
connect_heavy_edge(son[u], ancestor);
}
for(int i=head[u]; i; i=e[i].nxt)
{
int v = e[i].to;
if(v == son[u] || v == fa[u]) continue;
connect_heavy_edge(v, v);
}
}
void upd_ans(int x)
{
while(top[x] != top[1])
{
t.update(1, 1, n, dfn[top[x]], dfn[x], 1);
x = fa[top[x]];
}
t.update(1, 1, n, dfn[1], dfn[x], 1);
}
int get_ans(int x)
{
int ans = 0;
while(top[x] != top[1])
{
add(ans, t.query(1, 1, n, dfn[top[x]], dfn[x]));
x = fa[top[x]];
}
add(ans, t.query(1, 1, n, dfn[1], dfn[x]));
return ans;
}
void solve()
{
for(int i=1; i<=n; i++)
{
upd_ans(i);
for(que x : v[i]) ans[x.id] = (ans[x.id]+1ll*x.opt*get_ans(x.z)%mod+mod)%mod;
}
}
int main()
{
n = read(), m = read();
for(int i=2; i<=n; i++)
{
int x = read()+1; add_e(x, i);
}
find_heavy_edge(1, 0, 1);
connect_heavy_edge(1, 1);
for(int i=1; i<=m; i++)
{
int l = read(), r = read(), z = read();
v[l].push_back((que){i, -1, z+1});
v[r+1].push_back((que){i, 1, z+1});
}
solve();
for(int i=1; i<=m; i++) printf("%d\n", ans[i]);
return 0;
}
时光花火，水月星辰
