【AtCoder】AGC017
A - Biscuits
dp[i][0/1]表示当前和是偶数还是奇数，直接转移即可
#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 1000005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
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,P;
int a[55];
int64 dp[55][2];
void Solve() {
read(N);read(P);
for(int i = 1 ; i <= N ; ++i) read(a[i]);
dp[0][0] = 1;
for(int i = 1 ; i <= N ; ++i) {
int k = a[i] & 1;
dp[i][0] = dp[i - 1][0];dp[i][1] = dp[i - 1][1];
dp[i][k ^ 0] += dp[i - 1][0];
dp[i][k ^ 1] += dp[i - 1][1];
}
out(dp[N][P]);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
B - Moderate Differences
A和B差值在\(N \times C\)和\((N-1) \times D\)之间肯定能达到
因为下降操作和上升操作差不多，那么我们默认所有的下降操作都在前面
容易发现，当我们进行\(i\)次至少为\(C\)的下降后
\(-i \times C + (N - 1 - i) \times C\)和\(-i \times C + (N - 1 - i) \times D\)之间的也可以达到
且这样能构造最多的重合的区间
#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 1000005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
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;
int64 A,B,C,D;
bool check(int64 a,int64 b) {
return B - A >= a && B - A <= b;
}
void Solve() {
read(N);read(A);read(B);read(C);read(D);
for(int i = 0 ; i <= N - 1 ; ++i) {
int64 u = (N - 1 - i) * D - C * i,d = (N - 1 - i) * C - C * i;
if(check(d,u)) {puts("YES");return;}
u = C * i - (N - 1 - i) * C,d = C * i - (N - 1 - i) * D;
if(check(d,u)) {puts("YES");return;}
}
puts("NO");
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
C - Snuke and Spells
假如\(i\)出现了\(A[i]\)次，那么覆盖\([i - A[i],i]\)这个区间，然后找\([0,L]\)没有被覆盖的区间长度，就是答案
可以\(O(1)\)处理修改
#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 1000005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
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,M;
int A[MAXN];
int cnt[MAXN],cov[MAXN],ans;
void Solve() {
read(N);read(M);
for(int i = 1 ; i <= N ; ++i) {
read(A[i]);
cnt[A[i]]++;
cov[A[i] - cnt[A[i]] + 1]++;
}
for(int i = 1 ; i <= N ; ++i) {
if(!cov[i]) ++ans;
}
int x,y;
for(int i = 1 ; i <= M ; ++i) {
read(x);read(y);
if(A[x] - cnt[A[x]] + 1 >= 1) {
if(!--cov[A[x] - cnt[A[x]] + 1]) ++ans;
}
--cnt[A[x]];
A[x] = y;
++cnt[A[x]];
if(A[x] - cnt[A[x]] + 1 >= 1) {
if(!cov[A[x] - cnt[A[x]] + 1]++) --ans;
}
out(ans);enter;
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
D - Game on Tree
一个点的sg函数值显然是0，而一个树加一条边一个点的sg函数值是这棵树的sg函数值+1
证明，设新加的边为\(u,v\)，\(u\)是\(v\)的祖先，若断掉新加的边，则sg值是0
否则断树中的边，sg函数值为\([0,sg[v] - 1]\)，从小到大取，使得\(v\)为根游戏状态为0的那条边，断了之后，由于断掉新边游戏状态是0，则这个状态给\(u\)为根的贡献是\(1\)
使得\(v\)为根贡献为1的断边状态，子游戏中断边为0的状态可以转化为1，再填上断掉新边的状态是0，则这个对\(u\)的贡献变成了2
以此类推，这种情况游戏状态就是\(sg[v] + 1\)
#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 100005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
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;
struct node {
int to,next;
}E[MAXN * 2];
int sumE,sg[MAXN],head[MAXN];
void add(int u,int v) {
E[++sumE].to = v;
E[sumE].next = head[u];
head[u] = sumE;
}
void dfs(int u,int fa) {
sg[u] = 0;
for(int i = head[u]; i ; i = E[i].next) {
int v = E[i].to;
if(v != fa) {
dfs(v,u);
sg[u] ^= (sg[v] + 1);
}
}
}
void Solve() {
read(N);
int x,y;
for(int i = 1 ; i < N ; ++i) {
read(x);read(y);
add(x,y);add(y,x);
}
dfs(1,0);
if(sg[1]) puts("Alice");
else puts("Bob");
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
E - Jigsaw
把连在地面上的高度设为正数，不连在地面上的高度设为负数
然后一个点相当于一条边\((l,r)\)
相当于把整个图拆成负数点到正数点的若干路径
只要判断一个弱联通图，负数点是否全为度数是否全负，正数点度数是否全正，然后至少有一个点点度不为0
证明就是欧拉回路，两两匹配负数点和正数点，一定会有欧拉回路，断掉这些边就是答案
#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 10005
#define eps 1e-12
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
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 g[405][405],f[405][405],ind[405],H,col[405],all;
int N,A[100005],B[100005],C[100005],D[100005],cnt;
bool vis[405],flag,has[405];
void dfs(int u) {
all += ind[u];
if(ind[u]) flag = 1;
vis[u] = 1;
for(int i = 1 ; i <= 2 * H ; ++i) {
if(f[u][i] && !vis[i]) dfs(i);
}
}
void Solve() {
read(N);read(H);
for(int i = 1 ; i <= N ; ++i) {
read(A[i]);read(B[i]);read(C[i]);read(D[i]);
int s,t;
if(C[i]) s = C[i];
else s = A[i] + H;
if(D[i]) t = D[i] + H;
else t = B[i];
g[s][t]++;ind[t]++;ind[s]--;
f[s][t]++;f[t][s]++;
has[s] = has[t] = 1;
}
for(int i = 1 ; i <= H ; ++i) {
if(ind[i] < 0) {puts("NO");return;}
}
for(int i = H + 1 ; i <= 2 * H ; ++i) {
if(ind[i] > 0) {puts("NO");return;}
}
for(int i = 1 ; i <= 2 * H ; ++i) {
if(!has[i]) continue;
if(!vis[i]) {
flag = 0;
dfs(i);
if(all != 0 || !flag) {puts("NO");return;}
}
}
puts("YES");
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
F - Zigzag
dp[i][j][mask]表示第i条边走了第j步，左边界是什么
根据限制和每一步的选择修改左边界即可
复杂度\(O(n^{2}2^{n})\)
#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 eps 1e-12
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
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 = 1000000007;
int N,M,K;
int A[405],B[405],C[405],w[25][25];
int dp[2][21][(1 << 19) + 5];
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);
}
int lowbit(int x) {
return x & (-x);
}
void Solve() {
read(N);read(M);read(K);
memset(w,-1,sizeof(w));
for(int i = 1 ; i <= K ; ++i) {
read(A[i]);read(B[i]);read(C[i]);
w[A[i]][B[i]] = C[i];
}
int cur = 0;
dp[0][1][0] = 1;
for(int i = 1 ; i <= M ; ++i) {
for(int j = 1 ; j < N ; ++j) {
for(int s = 0 ; s < (1 << N - 1) ; ++s) {
if(!dp[cur][j][s]) continue;
if(w[i][j] != 1 && !(s >> (j - 1) & 1)) update(dp[cur][j + 1][s],dp[cur][j][s]);
if(w[i][j] != 0) {
if(s >> (j - 1) & 1) update(dp[cur][j + 1][s],dp[cur][j][s]);
else {
int t = s - (s & (1 << j) - 1);
t -= t & (-t);
t ^= (1 << j - 1);
t += s & (1 << j) - 1;
update(dp[cur][j + 1][t],dp[cur][j][s]);
}
}
}
}
memset(dp[cur ^ 1],0,sizeof(dp[cur ^ 1]));
for(int s = 0 ; s < (1 << N - 1) ; ++s) dp[cur ^ 1][1][s] = dp[cur][N][s];
cur ^= 1;
}
int ans = 0;
for(int s = 0 ; s < (1 << N - 1) ; ++s) ans = inc(ans,dp[cur][1][s]);
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
