XIII Open Grodno SU Championship

A. Alice in the Wonderland

按题意模拟。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
 
struct A
{
    int x, y, z;
}t, tt, st, ed;
queue<A> q;
int n, m, h;
char s[60][60][60];
bool e[60][60][60];
const int dx[4] = {0, 0, 1, -1}, dy[4] = {1, -1, 0, 0};
 
int main()
{
	scanf("%d%d%d", &n, &m, &h);
	for(int i = 1; i <= h; i ++){
        for(int j = 1; j <= n; j ++){
            scanf("%s", s[i][j] + 1);
        }
	}
	for(int i = 1; i <= h; i ++){
        for(int j = 1; j <= n; j ++){
            for(int k = 1; k <= m; k ++){
                if(s[i][j][k] == 'A'){
                    st.x = j; st.y = k; st.z = i;
                }
                else if(s[i][j][k] == 'E'){
                    ed.x = j; ed.y = k; ed.z = i;
                }
            }
        }
	}
	int flag = 0;
	q.push(st);
	e[st.z][st.x][st.y] = 1;
    while(! q.empty()){
        t = q.front(); q.pop();
        if(s[t.z][t.x][t.y] == 'E'){
            flag = 1;
            break;
        }
        if(s[t.z][t.x][t.y] == 'w'){
            int i;
            for(i = t.z; i <= h; i ++){ // 这里的方向要确认一下
                if(s[i][t.x][t.y] != 'w'){
                    break;
                }
            } i --;
            if(e[i][t.x][t.y] == 0){
                tt.z = i; tt.x = t.x; tt.y = t.y;
                q.push(tt);
                e[i][t.x][t.y] = 1;
            }
            if(i != t.z) continue;
        }
        if(s[t.z][t.x][t.y] == 's'){
            for(int i = t.z; i <= h; i ++){
                if(s[i][t.x][t.y] == 's' && e[i][t.x][t.y] == 0){
                    e[i][t.x][t.y] = 1;
                    tt.z = i; tt.x = t.x; tt.y = t.y;
                    q.push(tt);
                }
                else if(s[i][t.x][t.y] != 's') break;
            }
            for(int i = t.z; i >= 1; i --){
                if(s[i][t.x][t.y] == 's' && e[i][t.x][t.y] == 0){
                    e[i][t.x][t.y] = 1;
                    tt.z = i; tt.x = t.x; tt.y = t.y;
                    q.push(tt);
                }
                else if(s[i][t.x][t.y] != 's') break;
            }
 
        }
        for(int i = 0; i < 4; i ++){
            tt.x = t.x + dx[i];
            tt.y = t.y + dy[i];
            tt.z = t.z;
            if(tt.x >= 1 && tt.x <= n && tt.y >= 1 && tt.y <= m && e[tt.z][tt.x][tt.y] == 0){
                q.push(tt);
                e[tt.z][tt.x][tt.y] = 1;
            }
        }
    }
    if(flag) puts("Yes"); else puts("No");
 
	return 0;
}
/*
【trick&&吐槽】
 
【题意】
 
【分析】
 
3 3 3
A..
.w.
...
...
.wE
...
...
.w.
...
 
【时间复杂度&&优化】
3 3 3
...
s.E
...
...
s..
...
...
s.A
...
 
*/

　　

B. Batrachomyomachia

贪心，每次选承受能力最小的可行的。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 210, M = 1e4 + 10, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n, w;
const double eps = 1e-12;
int a[N][N], b[M];
double f[N][N];
bool del[111111];
multiset<int> sot;
multiset<int> :: iterator it, ir;
int sgn(double x){
    if(fabs(x) < eps) return 0;
    return x > 0 ? 1 : -1;
}
int main()
{
    scanf("%d%d", &n, &w);
    for(int i = 1; i < n; i ++) scanf("%d", &b[i]);
    sort(b+1,b+n);
    for(int i = 1; i <= 200; i ++){
        for(int j = 1; j <= i; j ++){
            a[i][j] = w;
        }
    }
    for(int i = 1; i <= 200; i ++){
        for(int j = 1; j <= i; j ++){
            f[i][j] = (f[i - 1][j] + f[i - 1][j - 1]) / 2 + (a[i - 1][j] + a[i - 1][j - 1]) / 2;
        }
    }
    for(int i=1;i<=200;i++){
        sort(f[i] + 1, f[i] + i + 1);
        reverse(f[i] + 1, f[i] + i + 1);
    }
    /*
    for(int i = 1; i <= 10; i ++){
        for(int j = 1; j <= i; j ++){
            printf("%.0f ", f[i][j]);
        }puts("");
    }
    */
    int ans = 0;
    for(int i = 2; i <= 200; i ++){
        for(int j = 1; j <= i; j ++){
            int flag=0;
            for(int k=1;k<n;k++)if(!del[k]&&sgn(b[k] - f[i][j])>=0){
                flag=k;
                break;
            }
 
            if(!flag){
                ans = i - 1;
                break;
            }
            del[flag]=1;
        }
        if(ans) break;
    }
    printf("%d\n", ans);
	return 0;
}
/*
【trick&&吐槽】
 
【题意】
 
【分析】
 
【时间复杂度&&优化】
 
*/

　　

C. Cherries

将所有数字排序，那么一定是选取连续$B-A+1$个数进行配对，枚举所有方案即可。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 5050, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n;
int a[N];
int main()
{
	while(~scanf("%d", &n))
	{
	    int A, B;
	    scanf("%d%d", &A, &B);
        for(int i = 1; i <= n; ++i)
        {
            scanf("%d", &a[i]);
        }
        sort(a + 1, a + n + 1);
        int len = B - A;
        LL ans = 1e18;
        for(int i = 1; i + len <= n; ++i)
        {
            int j = i + len;
            LL tmp = 0;
            for(int k = i, x = A; k <= j; ++k, ++x)
            {
                tmp += abs(a[k] - x);
            }
            gmin(ans, tmp);
        }
        printf("%lld\n", ans);
	}
 
	return 0;
}
/*
【trick&&吐槽】
 
【题意】
 
【分析】
 
【时间复杂度&&优化】
 
*/

　　

D. Divisibility Game

预处理出约数集合后爆搜+卡时即可通过。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
const int N=401;
const int lim=7;
const int lim2=3;
int i,j,x,now,ans,sum,ave,n,a[N],q[N];
vector<int>d[N];
int ED1 = CLOCKS_PER_SEC * 0.65;
int ED2 = CLOCKS_PER_SEC * 0.95;
void dfs(int s,int x){
 
    if(x>n){
        for(int i=1;i<=n;i++)printf("%d ",q[i]);
        exit(0);
    }
    if(clock() > ED1)
    {
        return;
        //puts("-1");
        //exit(0);
    }
    for(vector<int>::iterator w=d[s].begin();w!=d[s].end();w++)
        for(int j=min(a[*w],1);j;j--){
            a[*w]-=j;
            for(int o=0;o<j;o++)q[x+o]=*w;
            dfs(s+(*w)*j,x+j);
            a[*w]+=j;
        }
}
void dfs2(int s,int x){
 
    if(x>n){
        for(int i=1;i<=n;i++)printf("%d ",q[i]);
        exit(0);
    }
    if(clock() > ED2)
    {
        puts("-1");
        exit(0);
    }
    for(vector<int>::iterator w=d[s].begin();w!=d[s].end();w++)
        for(int j=min(a[*w],1);j;j--){
            a[*w]-=j;
            for(int o=0;o<j;o++)q[x+o]=*w;
            dfs2(s+(*w)*j,x+j);
            a[*w]+=j;
        }
}
int main(){
    scanf("%d",&n);
    for(i=0;i<n;i++)scanf("%d",&x),a[x]++,sum+=x;
    for(i=0;i<=sum;i++){
        for(j=lim+1;j<=13;j++)
     //   for(j=13;j;j--)
            if(i%j==0&&a[j])d[i].push_back(j);
        for(j=lim2+1;j<=lim;j++)
     //   for(j=13;j;j--)
            if(i%j==0&&a[j])d[i].push_back(j);
        for(j=lim2;j;j--)
            if(i%j==0&&a[j])d[i].push_back(j);
    }
    dfs(0,1);
    for(i=0;i<=sum;i++)reverse(d[i].begin(),d[i].end());
    dfs2(0,1);
    puts("-1");
}

　　

E. Enter the Word

设$dp[i]$表示打出前$i$个字符的最小代价，那么有$dp[i-1]\leq dp[i]\leq dp[i-1]+1$。

为了检查是否可以不$+1$，找到$dp$值的分界线，那么只要后面部分是前面部分的子串即可。

设$f[i]$表示子串匹配结束位置是$i$是否可行，可以通过bitset加速。

时间复杂度$O(\frac{n^2}{64})$。

#include<cstdio>
#include<bitset>
#include<cstring>
using namespace std;
const int N=200010;
int n,i,r,ans;char a[N];bitset<N>f,v[26];
int main(){
    scanf("%s",a);
    n=strlen(a);
    for(i=0;i<n;i++){
        a[i]-='a';
        f=f<<1&v[a[i]];
        if(!f.any()){
            for(int k=r;k<i;k++)v[a[k]][k]=1;
            r=i;
            ans++;
            f=v[a[i]];
        }
    }
    printf("%d",ans);
}

　　

F. Formula 1

按题意模拟即可，记录每个人的排名以及领先的圈数。

#include<cstdio>
#include<algorithm>
using namespace std;
const int N=100010;
int n,m,i,x,f[N],pos[N],g[N];
bool cmp(int x,int y){return g[x]==g[y]?pos[x]<pos[y]:g[x]>g[y];}
int main(){
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++){
        f[i]=i;
        pos[i]=i;
    }
    while(m--){
        scanf("%d",&x);
        for(i=1;i<=n;i++)if(f[i]==x)break;
        int o=i;
        if(o==1){
            g[x]++;
            for(i=1;i<n;i++)f[i]=f[i+1];
            f[n]=x;
            swap(f[n],f[n-1]);
        }else swap(f[o],f[o-1]);
        //for(i=1;i<=n;i++)printf("%d ",f[i]);puts("");
    }
    for(i=1;i<=n;i++)pos[f[i]]=i;
    sort(f+1,f+n+1,cmp);
    for(i=1;i<=n&&i<=6;i++)printf("%d ",f[i]);
}

　　

G. Game with Coins

将过程倒过来，则变成对于最后一个数，找到倒数第二个数，然后中间的数都要被最后一个数覆盖掉，区间DP即可。

#include<cstdio>
#include<algorithm>
using namespace std;
const int N=110;
int n,i,j,a[N],f[N][N],ans;
int dfs(int l,int r){
    if(~f[l][r])return f[l][r];
    int&t=f[l][r];
    t=0;
    int left=l+1,right=r;
    if(left>right)right+=n;
    for(int i=left;i<=right;i++)t=max(t,dfs(l,(i-1+n)%n)+dfs(i%n,r)+abs(a[l]-a[i%n]));
    return t;
}
int main(){
    scanf("%d",&n);
    for(i=0;i<n;i++)scanf("%d",&a[i]);
    for(i=0;i<n;i++)for(j=0;j<n;j++)if(i!=j)f[i][j]=-1;
    for(i=0;i<n;i++)for(j=0;j<n;j++)ans=max(ans,dfs(i,j));
    printf("%d",ans);
}

　　

H. Hamnattan

首先特判起点终点都在同一条街道上的情况。其余情况Dijkstra求最短路即可，需要现算代价。

#include<cstdio>
#include<queue>
#include<cstdlib>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
typedef pair<int,int>P;
typedef pair<ll,P>PI;
typedef pair<ll,PI>PII;
const ll inf=1LL<<60;
const int N=110,M=500000;
int sa[N],sb[N];
int n,m,i,j,x,y,a[N],b[N],ns[N][N],ew[N][N],s[N][N];
int sx,sy,ex,ey;
ll d[N][N];
int g[N][N],v[M][2],w[M][2],nxt[M],ed;
priority_queue<PI,vector<PI>,greater<PI> >q;
ll ans=inf;
inline void add(int x,int y,int xx,int yy,int z,int zz){
    v[++ed][0]=xx;
    v[ed][1]=yy;
    w[ed][0]=z;
    w[ed][1]=zz;
    nxt[ed]=g[x][y];
    g[x][y]=ed;
}
inline void add2(int x,int y,int xx,int yy,int w,int ww){
    add(x,y,xx,yy,w,ww);
    add(xx,yy,x,y,w,ww);
}
inline int col(int x,int y,ll z){
    int NS=ns[x][y],EW=ew[x][y],S=s[x][y];
    z%=NS+EW;
    if(S)return z<EW;
    return z>=NS;
}
inline ll cal(int x,int y,int dir,ll z){
    while(col(x,y,z)!=dir)z++;
    return z;
}
inline void ext(int x,int y,ll z){
    if(d[x][y]>z)q.push(PI(d[x][y]=z,P(x,y)));
}
void CHECK(){
    int i,j;
    for(i=1;i<=n;i++)for(j=1;j<=m;j++){
        if(sy==ey)if(i<n)if(sb[j-1]==sy)if(sa[i-1]<=sx&&sx<=sa[i])
        if(sa[i-1]<=ex&&ex<=sa[i]){
            printf("%d",abs(sx-ex)+abs(sy-ey));
            exit(0);
        }
        if(sx==ex)if(j<m)if(sa[i-1]==sx)if(sb[j-1]<=sy&&sy<=sb[j])
            if(sb[j-1]<=ey&&ey<=sb[j]){
            printf("%d",abs(sx-ex)+abs(sy-ey));
            exit(0);
        }
    }
}
void EXT(int x,int y){
    int i,j;
    for(i=1;i<=n;i++)for(j=1;j<=m;j++){
        if(i<n)if(sb[j-1]==y)if(sa[i-1]<=x&&x<=sa[i]){
            ext(i,j,cal(i,j,1,x-sa[i-1]));
            ext(i+1,j,cal(i+1,j,1,sa[i]-x));
        }
        if(j<m)if(sa[i-1]==x)if(sb[j-1]<=y&&y<=sb[j]){
            ext(i,j,cal(i,j,0,y-sb[j-1]));
            ext(i,j+1,cal(i,j+1,0,sb[j]-y));
        }
    }
}
void FIN(int x,int y){
    int i,j;
    for(i=1;i<=n;i++)for(j=1;j<=m;j++){
        if(i<n)if(sb[j-1]==y)if(sa[i-1]<=x&&x<=sa[i]){
            ans=min(ans,d[i][j]+x-sa[i-1]);
            ans=min(ans,d[i+1][j]+sa[i]-x);
        }
        if(j<m)if(sa[i-1]==x)if(sb[j-1]<=y&&y<=sb[j]){
            ans=min(ans,d[i][j]+y-sb[j-1]);
            ans=min(ans,d[i][j+1]+sb[j]-y);
        }
    }
}
int main(){
    scanf("%d%d",&n,&m);
    for(i=1;i<n;i++){
        scanf("%d",&a[i]);
        sa[i]=sa[i-1]+a[i];
    }
    for(i=1;i<m;i++){
        scanf("%d",&b[i]);
        sb[i]=sb[i-1]+b[i];
    }
    for(i=1;i<=n;i++)for(j=1;j<=m;j++){
        if(i<n)add2(i,j,i+1,j,a[i],1);
        if(j<m)add2(i,j,i,j+1,b[j],0);
    }
    for(j=1;j<=m;j++)for(i=1;i<=n;i++)scanf("%d%d%d",&ns[i][j],&ew[i][j],&s[i][j]);
    scanf("%d%d%d%d",&sx,&sy,&ex,&ey);
    CHECK();
    for(i=1;i<=n;i++)for(j=1;j<=m;j++)d[i][j]=inf;
    EXT(sx,sy);
    while(!q.empty()){
        PI t=q.top();q.pop();
        int x=t.second.first,y=t.second.second;
        if(d[x][y]<t.first)continue;
        //printf("%d %d %lld\n",x,y,t.first);
        for(i=g[x][y];i;i=nxt[i]){
            ext(v[i][0],v[i][1],cal(v[i][0],v[i][1],w[i][1],t.first+w[i][0]));
        }
    }
    FIN(ex,ey);
    printf("%lld",ans);
}
/*
4 3
10 10 10
10 10
1 99 0
99 1 0
50 99 0
1 99 1
1 99 0
99 1 0
20 41 1
1 99 0
99 1 0
1 99 1
99 1 0
99 1 0
1 10
30 19
*/

　　

I. Integer Pairs

只要$a[j]<0$即合法。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n;
int main()
{
    while(~scanf("%d", &n))
	{
	    int neg = 0;
        for(int i = 1; i <= n; ++i)
        {
            int x; scanf("%d", &x);
            neg += x < 0;
        }
        LL ans = neg * (n - 1ll);
        printf("%lld\n", ans);
	}
 
	return 0;
}
/*
【trick&&吐槽】
 
【题意】
 
【分析】
 
【时间复杂度&&优化】
3
-1 -2 -3
 
*/

　　

J. Jedi Training

线段树维护$f[l][r]$表示对应区间内选择子序列的左端点奇偶性为$l$，右端点奇偶性为$r$时的最大和。

#include<cstdio>
#include<algorithm>
using namespace std;
#define rep(i) for(int i=0;i<2;i++)
typedef long long ll;
const int N=100010,M=262150;
const ll inf=1LL<<60;
int n,m,i,a[N],op,x,y;
inline void up(ll&a,ll b){a<b?(a=b):0;}
struct E{
    ll f[2][2];
    E(){rep(i)rep(j)f[i][j]=-inf;}
    void clr(){rep(i)rep(j)f[i][j]=-inf;}
    void set(int x,ll p){
        clr();
        f[x&1][x&1]=p;
    }
    E operator+(const E&b){
        E c;
        rep(i)rep(j)c.f[i][j]=max(f[i][j],b.f[i][j]);
        rep(i)rep(j)rep(k)up(c.f[i][k],f[i][j]+b.f[j^1][k]);
        return c;
    }
    void write(){
        ll t=-inf;
        rep(i)rep(j)up(t,f[i][j]);
        printf("%lld\n",t);
    }
}v[M];
void build(int x,int a,int b){
    if(a==b){
        v[x].set(a,::a[a]);
        return;
    }
    int mid=(a+b)>>1;
    build(x<<1,a,mid),build(x<<1|1,mid+1,b);
    v[x]=v[x<<1]+v[x<<1|1];
}
void change(int x,int a,int b,int c,int d){
    if(a==b){
        v[x].set(a,d);
        return;
    }
    int mid=(a+b)>>1;
    if(c<=mid)change(x<<1,a,mid,c,d);else change(x<<1|1,mid+1,b,c,d);
    v[x]=v[x<<1]+v[x<<1|1];
}
E ask(int x,int a,int b,int c,int d){
    if(c<=a&&b<=d)return v[x];
    int mid=(a+b)>>1;
    E t;
    t.clr();
    if(c<=mid)t=ask(x<<1,a,mid,c,d);
    if(d>mid)t=t+ask(x<<1|1,mid+1,b,c,d);
    return t;
}
int main(){
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++)scanf("%d",&a[i]);
    build(1,1,n);
    while(m--){
        scanf("%d%d%d",&op,&x,&y);
        if(op==1)change(1,1,n,x,y);
        else{
            ask(1,1,n,x,y).write();
        }
    }
}

　　

K. Kings of a Round Table

假设$1$号国王一定位于$1$号位置，并不区分剩下$7$个国王，则这种情况下方案数还要乘以$n\times 7!$。

那么剩下的方案数大约只有$3\times 10^8$个，爆搜打表即可。

#include<cstdio>
long long f[111];
int n;
int main(){
  f[9]=0;
  f[10]=0;
  f[11]=0;
  f[12]=0;
  f[13]=0;
  f[14]=0;
  f[15]=0;
  f[16]=0;
  f[17]=685440;
  f[18]=725760;
  f[19]=11491200;
  f[20]=6451200;
  f[21]=83825280;
  f[22]=38142720;
  f[23]=397837440;
  f[24]=170311680;
  f[25]=1441440000;
  f[26]=617460480;
  f[27]=4330609920;
  f[28]=1905684480;
  f[29]=11330323200;
  f[30]=5175878400;
  f[31]=26645794560;
  f[32]=12675317760;
  f[33]=57564017280;
  f[34]=28504707840;
  f[35]=116035920000;
  f[36]=59698114560;
  f[37]=220799779200;
  f[38]=117723513600;
  f[39]=400156848000;
  f[40]=220502016000;
  f[41]=695520483840;
  f[42]=395054150400;
  f[43]=1165870379520;
  f[44]=680891904000;
  f[45]=1893253824000;
  f[46]=1134285546240;
  f[47]=2989486241280;
  f[48]=1833544581120;
  f[49]=4604213577600;
  f[50]=2885462496000;
  f[51]=6934509429120;
  f[52]=4433085296640;
  f[53]=10236190124160;
  f[54]=6664974140160;
  f[55]=14837041296000;
  f[56]=9826142699520;
  f[57]=21152159804160;
  f[58]=14230860215040;
  f[59]=29701625184000;
  f[60]=20277521510400;
  f[61]=41130725126400;
  f[62]=28465795572480;
  f[63]=56232969811200;
  f[64]=39416274616320;
  f[65]=75976140240000;
  f[66]=53892855878400;
  f[67]=101531626035840;
  f[68]=72828098703360;
  f[69]=134307318499200;
  f[70]=97351809811200;
  scanf("%d",&n);
  printf("%lld",f[n]);
}

　　

L. Lines and Polygon

求出直线与凸包的交点，然后在附近枚举即可得到最近点。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
 
const long double eps = 1e-8;
int sgn(double x)
{
    if(fabs(x) < eps) return 0;
    return x > 0 ? 1 : -1;
}
struct point
{
    long double x, y;
    point(){}
    point(long double a, long double b){x = a; y = b;}
    long double det (const point &b)const{
        return x * b.y - y * b.x;
    }
    friend bool operator < (const point &a, const point &b){
        if(sgn(a.x - b.x) == 0) return sgn(a.y - b.y) < 0;
        return sgn(a.x - b.x) < 0;
    }
    friend point operator - (const point &a, const point &b){
        return point(a.x - b.x, a.y - b.y);
    }
};
struct Convex
{
    int n;
    vector<point> a, upper, lower;
    Convex(){}
    Convex (vector<point> _a) : a(_a){
        n = a.size();
        int ptr = 0;
        for(int i = 1; i < n; i ++) if(a[ptr] < a[i]) ptr = i;
        for(int i = 0; i <= ptr; i ++) lower.push_back(a[i]);
        for(int i = ptr; i < n; i ++) upper.push_back(a[i]);
        upper.push_back(a[0]);
    }
    int sign(long double x){
        if(fabs(x) < eps) return 0;
        return x > 0 ? 1 : -1;
    }
 
    pair<long double, int> get_tangent(vector<point> &convex, point vec){
        int l = 0, r = (int) convex.size() - 2;
        for(; l + 1 < r;){
            int mid = (l + r) / 2;
            if(sign((convex[mid + 1] - convex[mid]).det(vec)) > 0) r = mid;
            else l = mid;
        }
        return max(make_pair(vec.det(convex[r]), r), make_pair(vec.det(convex[0]), 0));
    }
    int binary_search(point u, point v, int l, int r){
        int sl = sign((v - u).det(a[l % n] - u));
        for(; l + 1 < r;){
            int mid = (l + r) / 2;
            int smid = sign((v - u).det(a[mid % n] - u));
            if(smid == sl) l = mid;
            else r = mid;
        }
        return l % n;
    }
    int get_tangent(point vec){
        pair<long double, int> ret = get_tangent(upper, vec);
        ret.second = (ret.second + (int)lower.size() - 1) % n;
        ret = max(ret, get_tangent(lower, vec));
        return ret.second;
    }
    bool get_intersection(point u, point v, int &i0, int &i1){
        int p0 = get_tangent(u - v), p1 = get_tangent(v - u);
        if(sign((v - u).det(a[p0] - u)) * sign((v - u).det(a[p1] - u)) < 0){
            if(p0 > p1) swap(p0, p1);
            i0 = binary_search(u, v, p0, p1);
            i1 = binary_search(u, v, p1, p0 + n);
            return true;
        }
        else{
            return false;
        }
    }
};
int n;
point p[N];
vector<point> a, b;
Convex D;
int m;
long double A, B, C;
 
long double cal(int i0)
{
    return fabs((A * D.a[i0].x + B * D.a[i0].y + C) );
}
const double INF = 1e9;
int main()
{
	scanf("%d", &n);
	for(int i = 0; i < n; i ++) {
        //scanf("%lf%lf", &p[i].x, &p[i].y);
        double x, y;
        scanf("%lf%lf", &x, &y);
        p[i].x = x; p[i].y = y;
        //a.push_back(p[i]);
	}
	for(int i = n - 1; i >= 0; i --) a.push_back(p[i]);
	int ptr = 0;
	for(int i = 1; i < n; i ++){
        if(a[ptr] < a[i]) ptr = i;
	}
	for(int i = ptr; i < n; i ++){
        b.push_back(a[i]);
	}
	for(int i = 0; i < ptr; i ++){
        b.push_back(a[i]);
	}
	D = Convex(b);
	scanf("%d", &m);
    for(int i = 1; i <= m; i ++){
        double AA, BB, CC;
        scanf("%lf%lf%lf", &AA, &BB, &CC);
        A = AA; B = BB; C = CC;
        double ans = 1e18;
        int i0, i1;
        point p0, p1;
        if(A){
            p0.y = 0, p0.x = -C / A;
            p1.y = INF, p1.x = (- C - B * INF) / A;
        }
        else if(B){
            p0.x = 0, p0.y = -C / B;
            p1.x = INF, p1.y = (-C - INF * A) / B;
        }
        //else while(1);
        if(D.get_intersection(p0, p1, i0, i1)){
        #define next(i) ((i + 1) % n)
        #define pre(i) ((i - 1 + n) % n)
            for(int j = 0; j < 10; j ++){
                gmin(ans, cal(i0));
                i0 = next(i0);
            }
            for(int j = 0; j < 20; j ++){
                gmin(ans, cal(i0));
                i0 = pre(i0);
            }
            for(int j = 0; j < 10; j ++){
                gmin(ans, cal(i1));
                i1 = next(i1);
            }
            for(int j = 0; j < 20; j ++){
                gmin(ans, cal(i1));
                i1 = pre(i1);
            }
        }
        else{
            //ans = 0;
            //while(1);
            for(int j = 0; j < n; j ++) gmin(ans, cal(j));
        }
        double ANS = ans / sqrt(A * A + B * B);
        printf("%.6f\n", ANS);
    }
	return 0;
}
/*
【trick&&吐槽】
 
4
1 3
3 1
1 -1
-1 1
1
0 4 -5
【题意】
 
【分析】
 
【时间复杂度&&优化】
 
*/

　　

M. MIPT Campus

留坑。