【AtCoder】AGC024
A - Fairness
如果奇数次是b - a
否则是a - b
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
//#define ivorysi
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) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int64 A,B,C,K;
void Solve() {
read(A);read(B);read(C);read(K);
if(K & 1) {out(B - A);enter;}
else {out(A - B);enter;}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
B - Backfront
找数值最长的连续的一段子序列，然后将剩下的数必须要移动了
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 200005
//#define ivorysi
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) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int N,a[MAXN],pos[MAXN],dp[MAXN];
void Solve() {
read(N);
for(int i = 1 ; i <= N ; ++i) {
read(a[i]);
pos[a[i]] = i;
}
dp[1] = 1;
for(int i = 2 ; i <= N ; ++i) {
if(pos[i - 1] < pos[i]) dp[i] = dp[i - 1] + 1;
else dp[i] = 1;
}
int res = N;
for(int i = 1 ; i <= N ; ++i) {
res = min(res,N - dp[i]);
}
out(res);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
C - Sequence Growing Easy
数列肯定是一段连续递增的好几段拼起来，都找到加上最大值就可以了
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstring>
//#define ivorysi
#define pb push_back
#define MAXN 200005
#define eps 1e-12
#define space putchar(' ')
#define enter putchar('\n')
#define mp make_pair
#define fi first
#define se second
#define mo 974711
using namespace std;
typedef long long int64;
typedef double db;
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 - '0' + c;
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) putchar('-'),x = -x;
while(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int N,a[MAXN];
void Solve() {
read(N);
for(int i = 1 ; i <= N ; ++i) read(a[i]);
a[0] = -1;
for(int i = 1 ; i <= N ; ++i) {
if(a[i - 1] < a[i] - 1) {
puts("-1");
return;
}
}
int64 ans = 0;
for(int i = N ; i >= 1 ; --i) {
if(a[i] == a[i + 1] - 1) continue;
ans += a[i];
}
printf("%lld\n",ans);
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}
D - Isomorphism Freak
答案是直径的长度 / 2 + 1
我们对于直径长度为奇数，也就是有偶数个点，如果直径长度增加颜色也要增加，所以我们不增加直径
我们枚举作为直径中心的两个点，我们能达到这个要求仅当距离中心距离相同的点点度相同即可
对于直径长度为偶数，也就是有奇数个点，我们分两种情况，第一种是只有一个中心
第二种是有两个中心，我们枚举两个中心，然后要求这两个中心所在的树的深度最大的不超过中心的一半即可
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 105
//#define ivorysi
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) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
struct node {
int to,next;
}E[MAXN * 2];
int N,sumE,head[MAXN],A[MAXN],B[MAXN],dep[MAXN],ch[MAXN],x[MAXN];
void add(int u,int v) {
E[++sumE].to = v;E[sumE].next = head[u];head[u] = sumE;
}
int Calc(int u,int fa) {
int res = dep[u];
ch[u] = 0;
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(v != fa) {
ch[u]++;
dep[v] = dep[u] + 1;
res = max(res,Calc(v,u));
}
}
return res;
}
void Solve() {
read(N);
if(N == 1) {
puts("1 2");return;
}
for(int i = 1 ; i < N ; ++i) {
read(A[i]);read(B[i]);
add(A[i],B[i]);add(B[i],A[i]);
}
int D = 0;
for(int i = 1 ; i < N ; ++i) {
dep[A[i]] = dep[B[i]] = 0;
D = max(D,Calc(A[i],B[i]) + Calc(B[i],A[i]) + 1);
}
out(D / 2 + 1);space;
if(D & 1) {
int64 ans = 9e18;
for(int i = 1 ; i < N ; ++i) {
dep[A[i]] = dep[B[i]] = 0;
int64 tmp = 2;
if(Calc(A[i],B[i]) == D / 2 && Calc(B[i],A[i]) == D / 2) {
memset(x,0,sizeof(x));
for(int j = 1 ; j <= N ; ++j) x[dep[j]] = max(x[dep[j]],ch[j]);
for(int k = 0 ; k < D / 2 ; ++k) tmp = tmp * x[k];
ans = min(ans,tmp);
}
}
out(ans);enter;
}
else {
int64 ans = 9e18;
for(int i = 1 ; i < N ; ++i) {
dep[A[i]] = dep[B[i]] = 0;
int64 tmp = 1;
int x0 = Calc(A[i],B[i]),x1 = Calc(B[i],A[i]);
if(x0 > x1) {swap(x0,x1),swap(A[i],B[i]);}
if(x0 == D / 2 - 1 && x1 == D / 2) {
dep[B[i]] = 0;
Calc(B[i],0);
memset(x,0,sizeof(x));
for(int j = 1 ; j <= N ; ++j) x[dep[j]] = max(x[dep[j]],ch[j]);
for(int k = 0 ; k < D / 2 ; ++k) tmp = tmp * x[k];
ans = min(ans,tmp);
}
}
for(int i = 1 ; i < N ; ++i) {
dep[A[i]] = dep[B[i]] = 0;
int x0 = Calc(A[i],B[i]),x1 = Calc(B[i],A[i]);
if(max(x0,x1) <= D / 2) {
memset(x,0,sizeof(x));
int64 tmp = 2;
for(int j = 1 ; j <= N ; ++j) x[dep[j]] = max(x[dep[j]],ch[j]);
for(int k = 0 ; k <= D ; ++k) {
if(x[k]) tmp *= x[k];
}
ans = min(ans,tmp);
}
}
out(ans);enter;
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
E - Sequence Growing Hard
Let's restate the problem = =
我们就相当于先有一个一个标号为0，值为0的点
然后每次新加一个标号为k的点在p点上，我们要新加的k点严格大于p点的值
这个可以用dp来完成
dp[u][x]表示用u个点，根节点值为x的方案数，标号就是0 - u - 1
答案就是dp[N + 1][0]
转移就是
\(dp[n] = \sum_{y > x} dp[n - k][x] * dp[k][y] \binom{n - 2}{k - 1}\)
我们每次把y当做1号点，x当做0号点可以不重不漏
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
//#define ivorysi
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) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int N,K,MOD;
int dp[305][305],sum[305][305],C[305][305];
int inc(int a,int b) {
return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
return 1LL * a * b % MOD;
}
void update(int &x,int y) {
x = inc(x,y);
}
void Solve() {
read(N);read(K);read(MOD);
C[0][0] = 1;
for(int i = 1 ; i <= N ; ++i) {
C[i][0] = 1;
for(int j = 1 ; j <= i ; ++j) {
C[i][j] = inc(C[i - 1][j - 1],C[i - 1][j]);
}
}
for(int j = 0 ; j <= K ; ++j) {sum[1][j] = (j + 1) % MOD;dp[1][j] = 1;}
for(int i = 2 ; i <= N + 1; ++i) {
for(int j = 1 ; j < i ; ++j) {
for(int k = 0 ; k <= K ; ++k) {
update(dp[i][k],mul(mul(dp[j][k],inc(sum[i - j][K],MOD - sum[i - j][k])),C[i - 2][i - j - 1]));
}
}
sum[i][0] = dp[i][0];
for(int k = 1 ; k <= K ; ++k) sum[i][k] = inc(sum[i][k - 1],dp[i][k]);
}
out(dp[N + 1][0]);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
F - Simple Subsequence Problem
我们只要构造出一张图
例如110[00011001]
这个点可以转移到
1100[0011001]和1101[1001]和110[]
我们若s在集合中[s] = 1
每个点t的值是t[]
如果t是s的子序列，转移的路径是唯一的，所以只要用dp求路径条数即可
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define space putchar(' ')
#define enter putchar('\n')
#define mp make_pair
#define pb push_back
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
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){putchar('-');x = -x;}
if(x >= 10) out(x / 10);
putchar('0' + x % 10);
}
int N,K,cnt;
char s[25][(1 << 20) + 5];
vector<int> v[25];
int dp[42000005];
int p0[25],p1[25];
void Solve() {
read(N);read(K);
for(int i = 0 ; i <= N ; ++i) scanf("%s",s[i]);
for(int i = 1 ; i <= N ; ++i) {
v[i].resize((i + 1) * (1 << i));
int s = v[i].size();
for(int j = 0 ; j < s ; ++j) v[i][j] = ++cnt;
}
for(int i = 1 ; i <= N ; ++i) {
for(int j = 0 ; j < (1 << i) ; ++j) {
if(s[i][j] == '1') {
dp[v[i][j * (i + 1) + i]]++;
}
}
}
for(int i = N ; i >= 1 ; --i) {
for(int S = 0 ; S < (1 << i) ; ++S) {
p0[0] = -1,p1[0] = -1;
for(int j = 0 ; j <= i ; ++j) {
p1[j + 1] = p1[j];p0[j + 1] = p0[j];
if(S >> j & 1) p1[j + 1] = j;
else p0[j + 1] = j;
}
for(int j = i ; j >= 0 ; --j) {
int u = v[i][S * (i + 1) + j];
int b = S & ((1 << j) - 1),f = S >> j;
if(dp[u]) {
if(p0[j] != -1) {
int p = ((f << 1) << p0[j]) + (S & (1 << p0[j]) - 1);
int l = p0[j] + 1 + i - j;
dp[v[l][p * (l + 1) + p0[j]]] += dp[u];
}
if(p1[j] != -1) {
int p = ((f << 1 | 1) << p1[j]) + (S & (1 << p1[j]) - 1);
int l = p1[j] + 1 + i - j;
dp[v[l][p * (l + 1) + p1[j]]] += dp[u];
}
if(j != 0 && j != i) {
int l = i - j;
dp[v[l][f * (l + 1)]] += dp[u];
}
}
}
}
}
for(int i = N ; i >= 1 ; --i) {
for(int j = 0 ; j < (1 << i) ; ++j) {
if(dp[v[i][j * (i + 1)]] >= K) {
for(int k = i - 1; k >= 0 ; --k) {
putchar('0' + ((j >> k) & 1));
}
enter;
return;
}
}
}
puts("");
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
