【AtCoder】ARC102 题解
C - Triangular Relationship
题解
枚举一个数%K的值然后统计另两个
代码
#include <bits/stdc++.h>
#define enter putchar('\n')
#define space putchar(' ')
#define pii pair<int,int>
#define fi first
#define se second
#define MAXN 200005
#define pb push_back
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;T f = 1;char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) out(x / 10);
putchar('0' + x % 10);
}
int N,K;
int cnt[200005];
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
read(N);read(K);
int64 ans = 0;
for(int i = 1 ; i <= N ; ++i) {
cnt[i % K]++;
}
for(int i = 0 ; i < K ; ++i) {
if((K - i) * 2 % K == 0) {
ans += 1LL * cnt[i] * cnt[(K - i) % K] * cnt[(K - i) % K];
}
}
out(ans);enter;
return 0;
}
D - All Your Paths are Different Lengths
题解
感觉自己万分智障，没切掉，被学弟吊着打呀qwq
我们连出一条边全为\(2^0\)到\(2^(r - 1)\)的链出来，\(2^r\)是L的最高位
然后我们通过增加一些点到N的边权X，使得数量为L且合法即可。。
代码
#include <bits/stdc++.h>
#define enter putchar('\n')
#define space putchar(' ')
#define pii pair<int,int>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define eps 1e-8
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;T f = 1;char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 - '0' + c;
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) out(x / 10);
putchar('0' + x % 10);
}
int L,N;
void add(int u,int v,int c) {
out(u);space;out(v);space;out(c);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
read(L);
int r;
for(r = 20 ; r >= 0 ; --r) {
if(L & (1 << r)) break;
}
N = r + 1;out(N);space;int M = r * 2 + __builtin_popcount(L) - 1;out(M);enter;
for(int i = 1 ; i <= N - 1 ; ++i) {
add(i,i + 1,(1 << i - 1));add(i,i + 1,0);
}
for(int i = 1 ; i <= N - 1 ; ++i) {
if(L & (1 << i - 1)) {
L ^= (1 << i - 1);
add(i,N,L);
}
}
}
E - Stop. Otherwise...
题解
计数题
加和的限制相当于有几个数两个中只能出现一个
算一个dp[i][j]表示把i个数分成j段，每一段两种染色方式染色的方案数
然后枚举有多少数用来给两种方式染色，有多少数给怎么放都没事的数
如果有出现两次就不合法的数，那么就先算不存在它的排列，再算存在它一个的排列
代码
#include <bits/stdc++.h>
#define enter putchar('\n')
#define space putchar(' ')
#define pii pair<int,int>
#define fi first
#define se second
#define mp make_pair
#define MAXN 1000005
#define mo 999999137
#define pb push_back
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;T f = 1;char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) out(x / 10);
putchar('0' + x % 10);
}
const int MOD = 998244353;
int fac[4005],inv[4005],invfac[4005],N,K,vis[4005],pw[4005];
int dp[4005][4005];
int mul(int a,int b) {
return 1LL * a * b % MOD;
}
int inc(int a,int b) {
return a + b >= MOD ? a + b - MOD : a + b;
}
int C(int n,int m) {
if(n < m) return 0;
if(n < 0 || m < 0) return 0;
return mul(fac[n],mul(invfac[m],invfac[n - m]));
}
int fpow(int x,int c) {
int res = 1,t = x;
while(c) {
if(c & 1) res = mul(res,t);
t = mul(t,t);
c >>= 1;
}
return res;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
inv[1] = 1;pw[0] = 1;
for(int i = 2 ; i <= 4000 ; ++i) {
inv[i] = mul(inv[MOD % i],MOD - MOD / i);
}
fac[0] = invfac[0] = 1;
for(int i = 1 ; i <= 4000 ; ++i) {
fac[i] = mul(fac[i - 1],i);invfac[i] = mul(invfac[i - 1],inv[i]);
pw[i] = mul(pw[i - 1],2);
}
read(K);read(N);
dp[0][1] = 1;
for(int i = 1 ; i <= N ; ++i) dp[i][1] = 2;
dp[0][0] = 1;
for(int j = 2 ; j <= K ; ++j) {
int s = 1;
dp[0][j] = 1;
for(int i = 1 ; i <= N ; ++i) {
dp[i][j] = inc(mul(s,2),dp[i][j - 1]);
s = inc(s,dp[i][j - 1]);
}
}
for(int i = 2 ; i <= 2 * K ; ++i) {
int cnt = 0,t = 0,rest = 0;
memset(vis,0,sizeof(vis));
for(int j = 1 ; j <= K ; ++j) {
if(!vis[j] && j != i - j && i - j >= 1 && i - j <= K) {vis[j] = vis[i - j] = 1;++cnt;}
}
if(i % 2 == 0) ++t;
rest = K - cnt * 2 - t;
int ans = 0;
for(int j = N ; j >= 0 ; --j) {
ans = inc(ans,mul(dp[j][cnt],rest ? C(N - j - 1 + rest,rest - 1) : (N - j == 0)));
}
if(t) {
for(int j = N - 1; j >= 0 ; --j) {
ans = inc(ans,mul(dp[j][cnt],rest ? C(N - 2 - j + rest,rest - 1) : (N - 1 - j == 0)));
}
}
out(ans);enter;
}
return 0;
}
F - Revenge of BBuBBBlesort!
题解
我们按照逆操作考虑，容易发现是1-N顺序排列
每次交换\(a_{i - 1} < a_{i} < a_{i + 1}\)使得它变成排列p
这样原始排列每次交换的时候相邻两个不会都进行操作，所以排列p如果合法的话一定是
p[i] != i p[i + 1] == i + 1 p[i + 2] != i + 2....这样的排列
所以我们只要对p排列，顺序找出区间l,r保证l + 1,l + 3...r - 1是固定不动的
然后找出剩下往左往右的点，保证往左的点是顺序的，往右的点是顺序的即可
代码
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#define enter putchar('\n')
#define space putchar(' ')
#define fi first
#define se second
#define MAXN 300005
//#define ivorysi
#define pii pair<int,int>
using namespace std;
typedef long long int64;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {putchar('-');x = -x;}
if(x >= 10) out(x / 10);
putchar('0' + x % 10);
}
int N;
int p[MAXN],v[MAXN],cnt,id[MAXN],dir[MAXN];
bool vis[MAXN];
void Solve() {
read(N);
for(int i = 1 ; i <= N ; ++i) read(p[i]);
int la = 1;
while(1) {
while(p[la] == la && la <= N) ++la;
if(la > N) break;
int ed = la;
while(1) {
if(ed <= N && ((ed - la) & 1) && p[ed] == ed) ++ed;
else if(ed <= N && !((ed - la) & 1) && p[ed] != ed) ++ed;
else break;
}
--ed;
if(p[ed] == ed) --ed;
if(ed == la) {puts("No");return;}
cnt = 0;
for(int i = la ; i <= ed ; ++i) v[++cnt] = p[i];
sort(v + 1,v + cnt + 1);
for(int i = 1 ; i <= cnt ; ++i) {
if(v[i] != i + la - 1) {puts("No");return;}
}
cnt = 0;
for(int i = la ; i <= ed ; i += 2) {
v[++cnt] = p[i];
id[p[i]] = cnt;
}
sort(v + 1,v + cnt + 1);
for(int i = la ; i <= ed ; i += 2) {
int t = lower_bound(v + 1,v + cnt + 1,p[i]) - v;
if(t <= id[p[i]]) dir[i] = 0;
else dir[i] = 1;
}
int L = 0,R = 0;
for(int i = la ; i <= ed ; i += 2) {
if(dir[i] == 0) {
if(p[i] < L) {puts("No");return;}
L = p[i];
}
else {
if(p[i] < R) {puts("No");return;}
R = p[i];
}
}
la = ed + 1;
}
puts("Yes");
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}
