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2021CCPC威海
赛时过题与罚时
A.Goodbye, Ziyin!
签到题，队友写的
#include<bits/stdc++.h>
using namespace std;
int cnt[10], de[1000010];
int main() {
int n;
cin >> n;
for(int i = 1; i < n; ++i) {
int u, v;
scanf("%d %d", &u, &v);
++de[u];
++de[v];
}
for(int i = 1; i <= n; ++i) {
if(de[i] > 3) {
cout << "0";
return 0;
}
cnt[de[i]]++;
}
cout << cnt[1] + cnt[2];
return 0;
}
D.Period
队友写的KMP，签到题
#include <bits/stdc++.h>
using namespace std;
#define MAXN 1000005
char s[MAXN];
int nxt[MAXN];
int n;
int ans[MAXN];
int main()
{
scanf("%s", s + 1);
n = strlen(s + 1);
int j = 0;
for(int i = 2; i <= n; ++i)
{
while(j && s[i] != s[j + 1]) j = nxt[j];
if(s[i] == s[j + 1]) j++;
nxt[i] = j;
}
/* for(int i = 1; i <= n; ++i)
* {
* cout << nxt[i] << " ";
* }
* cout << "\n"; */
j = nxt[n];
while(j)
{
int l = j;
int r = n - l + 1;
if(r > l)
{
ans[r] --;
ans[l + 1] ++;
}
j = nxt[j];
}
for(int i = 1; i <= n; ++i)
ans[i] = ans[i] + ans[i-1];
/* for(int i = 1; i <= n; ++i)
* {
* cout << ans[i] << " ";
* }
* cout << "\n"; */
int q;
scanf("%d", &q);
while(q--)
{
int l;
scanf("%d", &l);
printf("%d\n", ans[l]);
}
return 0;
}
G. Shinyruo and KFC
简单组合计数
若
k
<
m
a
x
(
a
i
)
k<max(a_i)
k<max(ai)，显然无解
有解时答案是
∏
i
=
1
n
C
(
k
,
a
i
)
=
∏
i
=
1
n
k
!
a
i
!
∗
(
k
−
a
i
)
!
\prod_{i=1}^{n} C(k,a_i)=\prod_{i=1}^{n} \frac{k!}{{a_i}!*(k-a_i)!}
∏i=1nC(k,ai)=∏i=1nai!∗(k−ai)!k!
这样显然有一个
O
(
m
n
)
O(mn)
O(mn)的做法
观察数据性质
∑
i
=
1
n
a
i
<
=
1
e
5
\sum_{i=1}^{n}a_i<=1e5
∑i=1nai<=1e5
即
a
i
a_i
ai的种类数量级别为
s
q
r
t
(
n
)
sqrt(n)
sqrt(n)
那么就可以把相同的
a
i
a_i
ai一起做，优化成
O
(
m
n
)
O(m\sqrt{n})
O(mn)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5+5;
const int MOD = 998244353;
int n, m, a[N];
ll b[N];
ll c[N];
ll cnt[N];
ll power(ll x, ll y) {
if(y == 0) {
return 1;
}
ll res = 1;
while(y)
{
if(y&1) res = res * x % MOD;
x = x * x % MOD; y >>= 1;
}
return res;
}
ll ni(ll x) {
return power(x, MOD - 2);
}
int main() {
scanf("%d %d", &n, &m);
b[0] = 1;
for(ll i = 1; i <= 1e5; ++i) {
b[i] = b[i - 1] * i % MOD;
c[i] = power(b[i], n);
}
int mx = 0;
for(ll i = 1; i <= n; ++i ) {
scanf("%d", &a[i]);
++cnt[a[i]];
mx = max(mx, a[i]);
}
sort(a + 1, a + 1 + n);
n = unique(a + 1, a + 1 + n) - (a + 1);
for(int k = 1; k <= m; ++k) {
if(k < mx) {
puts("0");
continue;
}
ll ans = c[k];
for(ll i = 1; i <= n; ++i) {
ll t = b[a[i]] * b[k - a[i]] % MOD;
t = power(t, cnt[a[i]]);
ans = ans * ni(t) % MOD;
}
printf("%lld\n", ans);
}
return 0;
}
H. city safet
发现
n
n
n的范围很像网络流，然后就往这方面想，后面发现每当多选择一个距离相当于是获得了
v
i
+
1
−
v
i
v_{i+1}-v_{i}
vi+1−vi的贡献，每当修复了一个点，相当于是花费了
w
i
w_i
wi，我们要统一边权的性质，即我们可以想办法把问题变成最大贡献或者最小花费
那么我们可以假设初始时全部修复，
那么每个点建立
n
n
n个代表点，
i
d
i
,
j
id_{i,j}
idi,j表示以点
i
i
i为中心，距离
<
=
j
<=j
<=j的一类点，共
n
∗
n
n*n
n∗n个代表点
然后建立
n
n
n个具体点
1
1
1到
n
n
n
对于建图而言，
第
i
i
i个点的代表点
j
j
j向代表点
j
−
1
j-1
j−1连边，
(
i
,
j
)
−
>
(
i
,
j
−
1
)
(i,j)->(i,j-1)
(i,j)−>(i,j−1)，边权为
v
j
−
v
j
−
1
v_{j}-v_{j-1}
vj−vj−1，可理解成是反悔边，反悔这次距离的添加
每个代表点
(
i
,
j
)
(i,j)
(i,j)连向所有跟
i
i
i距离为
j
j
j的具体点，
(
i
,
j
)
−
>
k
当
d
i
s
[
i
]
[
k
]
=
j
(i,j)->k 当dis[i][k]=j
(i,j)−>k当dis[i][k]=j，边权为无穷
源点向所有代表点连边，边权为无穷，
S
−
>
(
i
,
j
)
S->(i,j)
S−>(i,j)
所有具体点向汇点连边，边权为对应的
w
i
w_i
wi，
i
−
>
T
i->T
i−>T，因为一个点最坏情况就是直接修了它
那么最小花费就是求最小割，直接网络流即可
那么答案就是
n
∗
v
n
−
m
a
x
f
l
o
w
n*v_n-maxflow
n∗vn−maxflow
#include<bits/stdc++.h>
#define INF 0x7fffffff
using namespace std;
const int N = 205;
const int M = 4e5+5;
struct node {
int to, next, w;
}e[M];
int n, m, k = 1, head[M], d[M];
int id[N][N], w[N], v[N];
int f[N][N];
int S, T, Maxflow, Num;
void build(int u, int v, int w) {
e[++k] = (node) {v, head[u], w};
head[u] = k;
e[++k] = (node) {u, head[v], 0};
head[v] = k;
}
queue<int> q;
bool bfs() {
while(!q.empty()) q.pop();
for(int i = S; i <= T; ++i) d[i] = 0;
d[S] = 1;
q.push(S);
while(!q.empty()) {
int u = q.front();
q.pop();
for(int i = head[u]; i; i = e[i].next){
int v = e[i].to;
if(e[i].w && !d[v]) {
d[v] = d[u] + 1;
q.push(v);
if(v == T) return 1;
}
}
}
return 0;
}
int dfs(int dep, int flow) {
if(dep == T) {
return flow;
}
int rest = flow, rp = 0;
for(int i = head[dep]; i && rest; i = e[i].next) {
int v = e[i].to;
if(d[v] == d[dep] + 1 && e[i].w) {
rp = dfs(v, min(e[i].w, rest));
if(!rp) {
d[v] = 0;
}
e[i].w -= rp;
e[i ^ 1].w += rp;
rest -= rp;
}
}
return flow - rest;
}
void dinic() {
int flow;
while(bfs()) {
while(flow = dfs(S, INF)) {
Maxflow += flow;
}
}
}
int main() {
scanf("%d", &n);
S = 0;
T = n*n+n+1;
for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
for (int i = 1; i <= n; i++) scanf("%d", &v[i]);
int U, V;
memset(f, 0x3f, sizeof(f));
for (int i = 1; i < n; i++) {
scanf("%d%d", &U, &V);
if (U != V) f[U][V] = f[V][U] = 1;
}
for (int i = 1; i <= n; i++) build(i, T, w[i]);
Num = n;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++) id[i][j] = ++Num;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
build(S, id[i][j], v[j]-v[j-1]);
build(id[i][j], id[i][j-1], INF);
}
}
for (int i = 1; i <= n; i++) f[i][i] = 0;
for (int k = 1; k <= n; k++)
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++) f[i][j] = min(f[i][j], f[i][k] + f[k][j]);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++) {
build(id[i][f[i][j]+1], j, INF);
}
dinic();
printf("%d\n", n*v[n]-Maxflow);
return 0;
}
J.Circular Billiard Table
签到题，我写的，题意是给一个入射角，求在圆内最少弹射多少次能回到原点
入射角是
a
b
\frac{a}{b}
ba，那么假设有
x
x
x个转角，要弹
y
y
y圈
显然有
a
b
∗
x
∗
2
=
360
∗
y
\frac{a}{b}*x*2=360*y
ba∗x∗2=360∗y
构造
x
=
180
∗
b
,
y
=
a
x=180*b,y=a
x=180∗b,y=a然后除
g
c
d
gcd
gcd就可以了
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {
return (!b) ? a : gcd(b, a%b);
}
ll t, a, b;
int main() {
scanf("%lld", &t);
while (t--) {
scanf("%lld%lld", &a, &b);
long long A = 180*b, B = a;
ll c = gcd(A, B);
ll ans = A/c-1;
printf("%lld\n", ans);
}
return 0;
}
M.810975
题意是构造一个长度为
n
n
n的01串，其中有
m
m
m个1，最长连续1 等于
k
k
k，问总方案数
相当于是在0的
n
−
m
+
1
n-m+1
n−m+1个空隙里面插入1
容斥一下，
答案即为
n
−
m
+
1
n-m+1
n−m+1个数中，满足
0
<
=
x
i
<
=
k
0<=x_i<=k
0<=xi<=k且
(
∑
i
=
1
n
−
m
+
1
x
i
)
=
m
(\sum_{i=1}^{n-m+1}x_i)=m
(∑i=1n−m+1xi)=m的方案数
减去
n
−
m
+
1
n-m+1
n−m+1个数，满足
0
<
=
x
i
<
=
k
−
1
0<=x_i<=k-1
0<=xi<=k−1且
(
∑
i
=
1
n
−
m
+
1
x
i
)
=
m
(\sum_{i=1}^{n-m+1}x_i)=m
(∑i=1n−m+1xi)=m的方案数
经典问题，队友直接秒了！
#include <bits/stdc++.h>
using namespace std;
#define mo 998244353
#define maxn 1000005
#define MAXN 1000005
#define ll long long
namespace Poly
{
inline void Add(int &x,int y){x+=y;x-=(x>=mo?mo:0);}
inline int add(int x,int y){x+=y;return x>=mo?x-mo:x;}
int A[maxn],B[maxn],C[maxn],D[maxn],E[maxn],inv[maxn],R[maxn],len,sz,ny;
inline int ksm(int x,int y=mo-2)
{
int a=1;if(y<0)y+=mo-1;
while(y)
{
if(y&1)a=1ll*a*x%mo;
x=1ll*x*x%mo;y>>=1;
}return a;
}
inline void init(int n)
{
int i;inv[1]=1;
for(i=2;i<=n;++i)inv[i]=1ll*(mo-mo/i)*inv[mo%i]%mo;
}
inline void Inte(int *a,int n)
{
for(int i=n-1;i>=1;--i)
a[i]=1ll*a[i-1]*inv[i]%mo;
a[0]=0;
}
inline void Deri(int *a,int n)
{
for(int i=1;i<n;++i)
a[i-1]=1ll*a[i]*i%mo;
a[n-1]=0;
}
inline void pre(int n)
{
len=1,sz=-1;while(len<n)len<<=1,++sz;ny=ksm(len);
for(int i=1;i<len;++i)R[i]=(R[i>>1]>>1)|((i&1)<<sz);
}
inline void NTT(int *a,int fl)//fl==1?3:332748118;
{
int i,j,k,t,y,p,w,wn;
for(i=1;i<len;++i)if(i<R[i])swap(a[i],a[R[i]]);
for(i=2,p=1;i<=len;p=i,i<<=1)
for(wn=ksm(fl,(mo-1)/i),j=0;j<len;j+=i)
for(w=1,k=0;k<p;++k,w=1ll*w*wn%mo)
{
t=a[j|k],y=1ll*w*a[j|k|p]%mo;
a[j|k]=add(t,y),a[j|k|p]=add(t,mo-y);
}
}
inline void Mul(int *a,int *b,int *c,int nm,int n,int m)//a is the ans;
{
if(!n||!m){fill(a,a+nm,0);return ;}pre(n+m-1);
for(int i=0;i<n;++i)A[i]=b[i];fill(A+n,A+len,0);
for(int i=0;i<m;++i)B[i]=c[i];fill(B+m,B+len,0);
NTT(A,3);NTT(B,3);
for(int i=0;i<len;++i)A[i]=1ll*A[i]*B[i]%mo;
NTT(A,332748118);
for(int i=0;i<nm;++i)a[i]=1ll*A[i]*ny%mo;
}
inline void Inv(int *a,int *b,int n)//this n include zero position;
{
if(n==1)return a[0]=ksm(b[0]),void();
int m=(n+1)>>1,i;Inv(a,b,m);pre(n<<1);
for(i=0;i<m;++i)A[i]=a[i];fill(A+m,A+len,0);
for(i=0;i<n;++i)B[i]=b[i];fill(B+n,B+len,0);
NTT(A,3);NTT(B,3);
for(i=0;i<len;++i)A[i]=1ll*A[i]*add(2,mo-1ll*A[i]*B[i]%mo)%mo;
NTT(A,332748118);
for(i=0;i<n;++i)a[i]=1ll*A[i]*ny%mo;
}
inline void Ln(int *a,int *b,int n)//a is the ans;b[0]need equal 1;
{
Inv(a,b,n);int i;pre(n<<1);
for(i=0;i<n;++i)C[i]=b[i];Deri(C,n);
for(i=0;i<n;++i)A[i]=a[i];fill(A+n,A+len,0);
for(i=0;i<n;++i)B[i]=C[i];fill(B+n,B+len,0);
NTT(A,3);NTT(B,3);
for(i=0;i<len;++i)A[i]=1ll*A[i]*B[i]%mo;
NTT(A,332748118);
for(i=0;i<n;++i)a[i]=1ll*A[i]*ny%mo;
Inte(a,n);
}
inline void Exp(int *a,int *b,int n)//a is the ans;
{
if(n==1)return a[0]=1,void();
int m=(n+1)>>1,i;Exp(a,b,m);Ln(D,a,n);pre(n<<1);
for(i=0;i<m;++i)A[i]=a[i];fill(A+m,A+len,0);
for(i=0;i<n;++i)B[i]=add(b[i],mo-D[i]);++B[0];fill(B+n,B+len,0);
NTT(A,3);NTT(B,3);
for(i=0;i<len;++i)A[i]=1ll*A[i]*B[i]%mo;
NTT(A,332748118);
for(i=0;i<n;++i)a[i]=1ll*A[i]*ny%mo;
}
inline void Ksm(int *a,int *b,int n, ll k)//a b should be differented;a is the ans;b[0] need equal 1;
{
Ln(E,b,n);k%=mo;
for(int i=0;i<n;++i)E[i]=1ll*E[i]*k%mo;
Exp(a,E,n);
}
}
int n, m, k;
int a[MAXN];
int b[MAXN];
int main()
{
cin >> n >> m >> k;
if(n == 0)
{
if(m == 0 && k == 0) { cout << 1 << "\n"; return 0; }
else { cout << 0 << "\n"; return 0; }
}
if(n < m) { cout << 0 << "\n"; return 0;}
if(m < k) { cout << 0 << "\n"; return 0; }
if(m == 0)
{
if(k == 0) { cout << 1 << "\n"; return 0;}
else { cout << 0 << "\n"; return 0;}
}
if(k == 0)
{
if(m == 0) { cout << 1 << "\n"; return 0;}
else { cout << 0 << "\n"; return 0; }
}
n = n - m + 1;
Poly::init(800000);
memset(a, 0, sizeof a);
for(int i = 0; i <= k; ++i) a[i] = 1;
Poly::Ksm(b, a, m + 1, n);
int ans = b[m];
memset(a, 0, sizeof a);
for(int i = 0; i < k; ++i) a[i] = 1;
memset(b, 0, sizeof b);
Poly::Ksm(b, a, m + 1, n);
ans = (ans - b[m] + mo) % mo;
cout << ans << "\n";
return 0;
}
