A - Biscuits
dp[i][0/1]表示當前和是偶數還是奇數，直接轉移即可
#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 ivorysifreopen("f1.in","r",stdin);
#endifSolve();
}B - Moderate Differences
A和B差值在\(N \times C\)和\((N-1) \times D\)之間肯定能達到
因為下降操作和上升操作差不多，那么我們默認所有的下降操作都在前面
容易發現，當我們進行\(i\)次至少為\(C\)的下降后
\(-i \times C + (N - 1 - i) \times C\)和\(-i \times C + (N - 1 - i) \times D\)之間的也可以達到
且這樣能構造最多的重合的區間
#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 ivorysifreopen("f1.in","r",stdin);
#endifSolve();
}C - Snuke and Spells
假如\(i\)出現了\(A[i]\)次，那么覆蓋\([i - A[i],i]\)這個區間，然后找\([0,L]\)沒有被覆蓋的區間長度，就是答案
可以\(O(1)\)處理修改
#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 ivorysifreopen("f1.in","r",stdin);
#endifSolve();
}D - Game on Tree
一個點的sg函數值顯然是0，而一個樹加一條邊一個點的sg函數值是這棵樹的sg函數值+1
證明，設新加的邊為\(u,v\)，\(u\)是\(v\)的祖先，若斷掉新加的邊，則sg值是0
否則斷樹中的邊，sg函數值為\([0,sg[v] - 1]\)，從小到大取，使得\(v\)為根游戲狀態為0的那條邊，斷了之后，由于斷掉新邊游戲狀態是0，則這個狀態給\(u\)為根的貢獻是\(1\)
使得\(v\)為根貢獻為1的斷邊狀態，子游戲中斷邊為0的狀態可以轉化為1，再填上斷掉新邊的狀態是0，則這個對\(u\)的貢獻變成了2
以此類推，這種情況游戲狀態就是\(sg[v] + 1\)
#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 ivorysifreopen("f1.in","r",stdin);
#endifSolve();
}E - Jigsaw
把連在地面上的高度設為正數，不連在地面上的高度設為負數
然后一個點相當于一條邊\((l,r)\)
相當于把整個圖拆成負數點到正數點的若干路徑
只要判斷一個弱聯通圖，負數點是否全為度數是否全負，正數點度數是否全正，然后至少有一個點點度不為0
證明就是歐拉回路，兩兩匹配負數點和正數點，一定會有歐拉回路，斷掉這些邊就是答案
#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 ivorysifreopen("f1.in","r",stdin);
#endifSolve();
}F - Zigzag
dp[i][j][mask]表示第i條邊走了第j步，左邊界是什么
根據限制和每一步的選擇修改左邊界即可
復雜度\(O(n^{2}2^{n})\)
#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 ivorysifreopen("f1.in","r",stdin);
#endifSolve();
}
 方法的重寫)
