AGC032
A - Limited Insertion
这题就是从后面找一个最靠后而且当前可以放的，可以放的条件是它的前面正好放了它的数值-1个数
如果不符合条件就退出
#include
#define fi first
#define se second
#define pii pair
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template
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
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int a[MAXN],N,cnt[MAXN];
bool vis[MAXN];
void Solve() {
read(N);
for(int i = 1 ; i <= N ; ++i) read(a[i]);
for(int i = 1 ; i <= N ; ++i) {
int cnt = 0;
for(int j = 1 ; j < i ; ++j) {
if(a[j] <= a[i]) ++cnt;
}
if(cnt + 1 < a[i]) {puts("-1");return;}
}
for(int i = 1 ; i <= N ; ++i) {
cnt[0] = 0;
for(int j = 1 ; j <= N ; ++j) cnt[j] = cnt[j - 1] + vis[j];
for(int j = N ; j >= 1 ; --j) {
if(!vis[j] && cnt[j - 1] + 1 == a[j]) {
vis[j] = 1;
out(a[j]);enter;
break;
}
}
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
} B - Balanced Neighbors
如果是偶数个，分成和相等的\(\frac{N}{2}\)份
如果是奇数个，最后一个点单独为一组，前\(N - 1\)个两个一组分成和相等的\(\frac{N - 1}{2}\)份
然后把一组作为一个点，建一个完全图，两组(一组1,2，一组3,4)之间连边就是
1-2,1-4,2-3,2-4
#include
#define fi first
#define se second
#define pii pair
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template
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
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int N;
vector v;
void Solve() {
read(N);
if(N & 1) {
for(int i = 1 ; i < N ; ++i) {
v.pb(mp(i,N));
}
--N;
}
for(int i = 1 ; i <= N ; ++i) {
for(int j = i + 1 ; j <= N ; ++j) {
if(i + j != N + 1) v.pb(mp(i,j));
}
}
out(v.size());enter;
for(auto t : v) {
out(t.fi);space;out(t.se);enter;
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
} C - Three Circuits
必定存在一个欧拉回路
如果一个点有六个或以上点度必定存在
如果有三个点以上有四个点度必定存在
如果只有两个点有四个点度
那么如果这四条路都在这两个点之间，那么就无解
否则有解
#include
#define fi first
#define se second
#define pii pair
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template
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
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 head[MAXN],sumE;
int N,M;
int cnt[MAXN];
bool vis[MAXN];
void add(int u,int v) {
E[++sumE].to = v;
E[sumE].next = head[u];
head[u] = sumE;
}
void dfs(int u) {
vis[u] = 1;
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(!vis[v]) dfs(v);
}
}
void Solve() {
read(N);read(M);
int a,b;
for(int i = 1 ; i <= M ; ++i) {
read(a);read(b);
cnt[a]++;cnt[b]++;
add(a,b);add(b,a);
}
for(int i = 1 ; i <= N ; ++i) {
if(cnt[i] & 1) {puts("No");return;}
}
int t = 0;
int p = 0,q = 0;
for(int i = 1 ; i <= N ; ++i) {
if(cnt[i] >= 6) {puts("Yes");return;}
if(cnt[i] >= 4) {
++t;
if(!p) p = i;
else if(!q) q = i;
}
}
if(t > 2) {puts("Yes");return;}
if(t == 2) {
vis[p] = 1;
dfs(q);
for(int i = 1 ; i <= N ; ++i) {
if(!vis[i]) {puts("Yes");return;}
}
}
puts("No");
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
} D - Rotation Sort
把操作改成数轴上，我可以把一个点移动到左边或右边任意一个位置上，可以不必要是整数点
显然对于每个点操作只进行一次
然后拆成\((-\infty,1),(1,2),(2,3),(3,4)....(N - 1,N),(N,+\infty)\)和整数点\(1,2,3,4,5,6..N\)
然后根据位置dp即可
#include
#define fi first
#define se second
#define pii pair
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 5005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template
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
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;
int64 dp[MAXN][2 * MAXN],s[MAXN][2 * MAXN];
int p[MAXN],pos[MAXN];
void Solve() {
read(N);read(A);read(B);
for(int i = 1 ; i <= N ; ++i) {read(p[i]);pos[p[i]] = i;}
for(int i = 1 ; i <= N ; ++i) {
s[i][0] = 1e18;
for(int j = 1 ; j <= 2 * N + 1 ; ++j) {
if(j & 1) {
dp[i][j] = s[i - 1][j] + (j < pos[i] * 2 ? B : A);
}
else {
dp[i][j] = s[i - 1][j - 1] + (j != pos[i] * 2 ? (j < pos[i] * 2 ? B : A): 0);
}
s[i][j] = min(s[i][j - 1],dp[i][j]);
}
}
out(s[N][2 * N + 1]);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
} E - Modulo Pairing
用蓝色表示相加小于M的，红色表示大于M的
然后就会变成前面是蓝的，后面是红的
简单分析发现蓝的越小越好，红的越大越好，所以求出能向左最长的红色序列
#include
#define fi first
#define se second
#define pii pair
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template
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
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];
void Solve() {
read(N);read(M);
for(int i = 1 ; i <= 2 * N ; ++i) read(a[i]);
sort(a + 1,a + 2 * N + 1);
int r = -1;
for(int i = 2 * N ; i >= 1 ; --i) {
int t = lower_bound(a + 1,a + 2 * N + 1,M - a[i]) - a;
if((t & 1) == (i & 1)) ++t;
if(r == -1) r = i + t;
else r = max(r,i + t);
}
if(r == -1) r = 4 * N + 1;
int ans = 0;
for(int i = 2 * N ; i >= 1 ; --i) {
if(r - i < i) ans = max(ans,(a[i] + a[r - i]) % M);
else break;
}
r = 1 + r - 2 * N - 1;
for(int i = 1 ; i <= 2 * N ; ++i) {
if(i < r - i) ans = max(ans,a[i] + a[r - i]);
else break;
}
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
} F - One Third
切一刀，画一条红线，然后正120画一条蓝线，反120画一条绿线
我们要求的就是任意1/3中最短两个颜色不同之间的线的角度大小
然后我们取第一刀红线和蓝线之间的 1/3区间，然后里面被分成了\(N\)份，我们递推一个\(f(i)\)，表示有\(i\)份左右两边区间颜色不同的概率
然后\(g(i)\)表示\(i\)份中最短的一段的期望长度，\(\frac{1}{3}\)长度里\(N\)份中选\(i\)份期望长度是\(\frac{i}{3N}\)，可以认为分成\(N\)份中选\(i\)份每个点被选中的概率是\(\frac{i}{N} = \frac{\binom{i - 1}{N - 1}}{\binom{i}{N}}\)
那\(i\)段中的期望最小长度呢
是\(E(min) = \int_{t = 0}^{1/i}P(min \geq t) dt = \int_{t = 0}^{1/i}(1 - it)^{i - 1}dt = \int_{t = 0}^{1}\frac{t^{i - 1}}{i} dt = \frac{1}{i^2}\)
所以\(g(i) = \frac{1}{3iN}\)
答案就是所有的\(f(i)g(i)\)的和
#include
#define fi first
#define se second
#define pii pair
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 1000005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template
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
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;
int f[MAXN][3],g[MAXN],fac[MAXN],invfac[MAXN],inv[MAXN];
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;
}
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 C(int n,int m) {
if(n < m) return 0;
return mul(fac[n],mul(invfac[m],invfac[n - m]));
}
void Solve() {
read(N);
fac[0] = 1;
for(int i = 1 ; i <= N; ++i) fac[i] = mul(fac[i - 1],i);
invfac[N] = fpow(fac[N],MOD - 2);
for(int i = N - 1 ; i >= 0 ; --i) invfac[i] = mul(invfac[i + 1],i + 1);
inv[1] = 1;
for(int i = 2 ; i <= 1000000 ; ++i) inv[i] = mul(inv[MOD % i],MOD - MOD / i);
f[0][0] = 1;
int iv3= fpow(inv[3],N - 1);
int ans = 0;
for(int i = 1 ; i <= N ; ++i) {
f[i][1] = inc(f[i - 1][0],f[i - 1][2]);
f[i][0] = inc(f[i - 1][1],f[i - 1][2]);
f[i][2] = inc(f[i - 1][0],f[i - 1][1]);
g[i] = mul(mul(inv[3],inv[N]),inv[i]);
int p = mul(mul(f[i][1],C(N,i)),iv3);
ans = inc(ans,mul(g[i],p));
}
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}