XIV Open Cup named after E.V. Pankratiev. GP of America

A. Ancient Diplomacy

建图,同色点间边权为$0$,异色点间边权为$1$,则等价于找一个点使得到它最短路最长的点的最短路最小,Floyd即可。

时间复杂度$O(n^3)$。

#include<cstdio>
#include<algorithm>
using namespace std;
const int N=110,inf=100000000;
int n,m,i,j,k,x,y,a[N],g[N][N];
int main(){
	while(~scanf("%d%d",&n,&m)){
		if(!n)return 0;
		for(i=1;i<=n;i++)scanf("%d",&a[i]);
		for(i=1;i<=n;i++)for(j=1;j<=n;j++)g[i][j]=i==j?0:inf;
		while(m--){
			scanf("%d%d",&x,&y);
			g[x][y]=g[y][x]=a[x]^a[y];
		}
		for(k=1;k<=n;k++)for(i=1;i<=n;i++)for(j=1;j<=n;j++)g[i][j]=min(g[i][j],g[i][k]+g[k][j]);
		int ans=inf;
		for(i=1;i<=n;i++){
			int now=0;
			for(j=1;j<=n;j++)now=max(now,g[i][j]);
			ans=min(ans,now);
		}
		printf("%d\n",ans);
	}
}

  

B. Bob and Banjo

若$AB$线段经过圆$C$的部分长度不超过$t$,则答案为$AB$长度。

否则若$t=0$,则答案为$A$到$C$切线长度$+B$到$C$切线长度$+$一段圆弧。

否则考虑最优路径,一定是$A$沿直线走到圆上某个点$D$,在圆内长度不超过$t$,然后从$D$开始走若干个长度为$t$的折线到达圆上某个点$E$,再由$E$沿直线到达$B$,中间不经过圆。

枚举顺时针还是逆时针走,再枚举$D$到$E$路径上有多少个$t$,那么可以解出$DE$这段圆心角的取值范围,在里面三分答案即可。

#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() {  }
#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 = 1e6 + 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 double eps = 1e-8, pi = acos(-1.0);
int sgn(double x)
{
    if(x < -eps) return -1;
    if(x > eps) return 1;
    return 0;
}
bool Quadratic(double A, double B, double C, double *t0, double *t1)
{
    double discrim = B * B - 4.f * A * C;
    if(discrim < 0.) return false;
    double rootDiscrim = sqrt(discrim);
    double q;
    if(B < 0) q = -.5f * (B - rootDiscrim);
    else      q = -.5f * (B + rootDiscrim);
    *t0 = q / A;
    *t1 = C / q;
    if(*t0 > *t1) swap(*t0, *t1);
    return true;
}
struct vec
{
    double x, y; vec(){x = y = 0;}
    vec(double _x, double _y) {x = _x, y = _y;}
    vec operator + (vec v){return vec(x + v.x, y + v.y);}
    vec operator - (vec v){return vec(x - v.x, y - v.y);}
    vec operator * (double v) {return vec(x * v, y * v);}
    vec operator / (double v) {return vec(x / v, y / v);}
    double operator * (vec v) {return x * v.x + y * v.y;}
    double len(){return hypot(x, y);}
    double len_sqr(){return x * x + y * y;}
    vec rotate(double c){
        return vec(x * cos(c) - y * sin(c), x * sin(c) + y * cos(c));
    }
    vec trunc(double l){return (*this) * l / len();}
    vec tot90(){return vec(-y, x);}
};
vec lerp(vec a, vec b, double t){return a * (1 - t) + b * t;}

struct circle
{
    vec c;
    double r;
    circle(){c = vec(0, 0), r = 0;}
    circle(vec _c, double _r){c = _c, r = _r;}
    vec point(double a){return vec(c.x + r * cos(a), c.y + r * sin(a));}
};

    circle c;
    vec st, ed;
    double t, ans,th;
int getTangets(vec p, circle C, double *v,double&cen,int mode=0)
{
    double x = atan2(p.y - C.c.y, p.x - C.c.x);
    double dist = (C.c - p).len();
    double ang = acos(C.r / dist);
    if(mode){
        ang=acos(sqrt(c.r*c.r-t*t/4)/dist);
        ang+=th/2;
    }
    v[0] = x + ang;  v[1] = x - ang;
    if(v[0]<v[1])swap(v[0],v[1]);
    cen=x;
    return 2;
}
bool circle_line_intersection(circle c, vec a, vec b, double *t0, double *t1)
{
    vec d = b - a;
    double A = d * d;
    double B = d * (a - c.c) * 2.0;
    double C = (a - c.c).len_sqr() - c.r * c.r;
    return Quadratic(A, B, C, t0, t1);
}

inline double cal(double x,double offset){
    return (st-c.point(x)).len()+(ed-c.point(x+offset)).len();
}
void gao(double v10,double v11,double v20,double v21){
    double lim1=v21-v10,lim2=v20-v11;
    //if(v10<v11)while(1);
    //if(v20<v21)while(1);
    //if(v10>v21)while(1);
    //if(lim1>lim2)while(1);
    lim1/=th;
    lim2/=th;
    //if(sgn(th)==0)while(1);
    int mx=((int)lim2)+10;
    int _l=max(1,(int)lim1-10),_r=mx;
    //if(lim1>lim2)while(1);
    for(int k = max(1,(int)lim1-10); k<=mx; k ++){
                    //v1[1]<=x<=v1[0]
                    //v2[1]-th<=x+k*th<=v2[0]
                    double L=max(v11,v21-k*th);
                    double R=min(v10,v20-k*th);
                    if(sgn(L-R)>0)continue;
                    for(int _=50;_--;){
                        double len=(R-L)/3;
                        double m1=L+len,m2=R-len;
                        double f1=cal(m1,k*th),f2=cal(m2,k*th);
                        if(f1<f2){
                            ans=min(ans,f1+k*t);
                            R=m2;
                        }else{
                            ans=min(ans,f2+k*t);
                            L=m1;
                        }
                        //if(L+1e-4>R)break;
                    }
                }
}
int main()
{
    while(~ scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &st.x, &st.y, &ed.x, &ed.y, &c.c.x, &c.c.y, &c.r, &t)){
        if(sgn(st.x) == 0 && sgn(st.y) == 0 && sgn(ed.x) == 0 && sgn(ed.y) == 0 && sgn(c.c.x) == 0 && sgn(c.c.y) == 0 && sgn(c.r) == 0 && sgn(t) == 0) break;
        double t0, t1;
        if(circle_line_intersection(c, st, ed, &t0, &t1)){ // 有交点
            vec inter1, inter2;
            if((sgn(t0)<=0||sgn(t0-1)>=0)&&(sgn(t1)<=0||sgn(t1-1)>=0)){
                    ans = (st - ed).len();
            }else{
            inter1 = lerp(st, ed, t0);
            inter2 = lerp(st, ed, t1);
            if(sgn((inter1 - inter2).len() - t) <= 0){
                ans = (st - ed).len();
            }
            else if(sgn(t)){
                ans=1e100;
                th = asin(t / 2 / c.r) * 2;
                for(int i=0;i<2;i++){
                    double v1[2],c1, v2[2],c2;
                    getTangets(st, c, v1,c1,1);
                    getTangets(ed, c, v2,c2);
                    for(int i=0;i<10;i++){
                        v2[0]-=pi*2;
                        v2[1]-=pi*2;
                        c2-=pi*2;
                    }
                    while(v2[1]<c1)v2[0]+=pi*2,v2[1]+=pi*2,c2+=pi*2;
                    th = asin(t / 2 / c.r) * 2;
                    //if(th<0)while(1);
                    gao(v1[0],v1[1],v2[0],v2[1]);
                    gao(v1[0],c1,c2,v2[1]);
                    gao(c1,v1[1],v2[0],c2);
                    swap(st,ed);
                }
                //gao(v1[0],c1,c2,v2[1]);
                //gao(c1,v1[1],v2[0],c2);
            }else{
                double v1[2],c1, v2[2],c2;
                getTangets(st, c, v1,c1);
                getTangets(ed, c, v2,c2);
                ans=1e100;
                for(int i=0;i<2;i++)for(int j=0;j<2;j++){
                        vec a=c.point(v1[i]);
                        vec b=c.point(v2[j]);
                        double now=(st-a).len()+(ed-b).len();
                        double o=fabs(v1[i]-v2[j]);
                        o=min(o,pi*2-o);
                        ans=min(ans,now+o*c.r);
                }
            }
            }
        }
        else{
            ans = (st - ed).len();
        }
        if(ans>1e20)while(1);
        printf("%.15f\n",ans);
    }
	return 0;
}

/*
【trick&&吐槽】
0 0 10 0 5 0 3 5


【题意】


【分析】


【时间复杂度&&优化】


*/

  

C. Chess Knight's Poem

设$v[i][a][b][c][d]$表示匹配了前$i$个字符,两个棋子分别在$(a,b)$和$(c,d)$是否可能,然后BFS即可。

#include<cstdio>
#include<cstring>
const int N=111;
int dx[8]={-2,-2,2,2,-1,-1,1,1};
int dy[8]={-1,1,-1,1,2,-2,2,-2};
char xiao[4][10]={
{'q','w','e','r','t','y','u','i','o','p'},
{'a','s','d','f','g','h','j','k','l',';'},
{'z','x','c','v','b','n','m',',','.','/'},
{0,0,1,1,1,1,1,1,0,0}
};
char da[4][10]={
{'Q','W','E','R','T','Y','U','I','O','P'},
{'A','S','D','F','G','H','J','K','L',':'},
{'Z','X','C','V','B','N','M','<','>','?'},
{0,0,1,1,1,1,1,1,0,0}
};
int n,i,j,k,x,y,z;
bool v[N][4][10][4][10];
int q[N*40*40][5],h,t;
bool flag;
char a[N];
inline void ext(int o,int A,int B,int C,int D,int _){
	if(A<0||A>3)return;
	if(B<0||B>9)return;
	if(C<0||C>3)return;
	if(D<0||D>9)return;
	if(A==C&&B==D)return;
	if(_==1){
		if(xiao[A][B]>1){//not shift not space
			if(xiao[C][D]==0){//shift
				if(a[o]!=da[A][B])return;
				o++;
			}else{
				if(a[o]!=xiao[A][B])return;
				o++;
			}
		}else if(xiao[A][B]==1){
			if(a[o]!=' ')return;
			o++;
		}
	}
	if(_==2){
		if(xiao[C][D]>1){//not shift not space
			if(xiao[A][B]==0){//shift
				if(a[o]!=da[C][D])return;
				o++;
			}else{
				if(a[o]!=xiao[C][D])return;
				o++;
			}
		}else if(xiao[C][D]==1){
			if(a[o]!=' ')return;
			o++;
		}
	}
	if(v[o][A][B][C][D])return;
	v[o][A][B][C][D]=1;
	q[++t][0]=o;
	q[t][1]=A;
	q[t][2]=B;
	q[t][3]=C;
	q[t][4]=D;
}
int main(){
	while(1){
		gets(a+1);
		n=strlen(a+1);
		if(a[1]=='*')return 0;
		for(i=0;i<=n+5;i++)for(j=0;j<4;j++)for(k=0;k<10;k++)for(x=0;x<4;x++)for(y=0;y<10;y++)v[i][j][k][x][y]=0;
		h=1,t=0;
		ext(1,3,0,3,9,0);
		flag=0;
		while(h<=t){
			x=q[h][0];
			if(x>n){
				flag=1;
				break;
			}
			y=q[h][1];
			z=q[h][2];
			int A=q[h][3];
			int B=q[h][4];
			h++;
			for(int i=0;i<8;i++){
				ext(x,y+dx[i],z+dy[i],A,B,1);
				ext(x,y,z,A+dx[i],B+dy[i],2);
			}
		}
		puts(flag?"1":"0");
	}
}
/*
CAlmimg eventa
*/

  

D. Drone

将$(t,f(t))$的函数画图,可以发现最大值和最小值都在凸壳上,因此答案是可以三分的。

#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() {  }
#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 = 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;
int n;
double x[N], v[N];
double check(double t)
{	
	double mx = -1e18;
	double mn= 1e18;
	for(int i = 1; i <= n; ++i)
	{
		double p = x[i] + t * v[i];
		gmax(mx, p);
		gmin(mn, p);
	}
	return mx - mn;
}
int main()
{
	while(~scanf("%d", &n), n)
	{
		for(int i = 1; i <= n; ++i)
		{
			scanf("%lf%lf", &x[i], &v[i]);
		}
		double l = 0;
		double r = 1.001e10;
		for(int tim = 1; tim <= 100; ++tim)
		{
			double lt = (l + l + r) / 3;
			double rt = (l + r + r) / 3;
			double lv = check(lt);
			double rv = check(rt);
			lv <= rv ? r = rt : l = lt;
		}
		printf("%.12f\n", check(l));
	}
	return 0;
}
/*
【trick&&吐槽】
2
-100 1
100 -1

3
-100 1
100 -1
101 -1

3
-100 -1
0 0
100 1

0

【题意】


【分析】


【时间复杂度&&优化】


*/

  

E. Ed and The Legend of Zelda

设$f[i][j][k][l]$表示考虑$i$的子树,$i$子树内作弊了$j$次,$i$作弊情况为$k$,$i$有$l$个儿子作弊的方案数,使用组合数转移。

时间复杂度$O(n^2)$。

#include<cstdio>
const int N=205,P=1000000007;
int n,m,i,x,g[N],nxt[N],f[N][N][2];
int size[N],ans;
int fac[N],inv[N];
int h[N][2][2];
int dp[N][2][2];
inline void up(int&a,int b){a=a+b<P?a+b:a+b-P;}
inline int C(int n,int m){return 1LL*fac[n]*inv[m]%P*inv[n-m]%P;}
void dfs(int x){
	size[x]=1;
	for(int i=g[x];i;i=nxt[i])dfs(i),size[x]+=size[i];
	for(int j=0;j<=1;j++)for(int k=0;k<2;k++)for(int o=0;o<2;o++)dp[j][k][o]=0;
	dp[0][0][0]=fac[size[x]-1];
	dp[1][1][0]=fac[size[x]-1];
	int pre=1;
	for(int i=g[x];i;i=nxt[i]){
		for(int j=0;j<=pre+size[i];j++)for(int k=0;k<2;k++)for(int o=0;o<2;o++)h[j][k][o]=0;
		for(int j=0;j<=pre;j++)for(int k=0;k<2;k++)for(int o=0;o<2;o++)if(dp[j][k][o]){
			for(int A=0;A<=size[i];A++)for(int B=0;B<2;B++)if(f[i][A][B]){
				if(k&&B)continue;
				if(o&&B)continue;
				if(B){
					up(h[j+A][k][o+B],1LL*dp[j][k][o]*inv[size[x]-1]%P*fac[size[x]-size[i]-1]%P*C(size[x]-1,size[i]-1)%P*f[i][A][B]%P);
				}else{
					up(h[j+A][k][o+B],1LL*dp[j][k][o]*inv[size[i]]%P*f[i][A][B]%P);
				}
			}
		}
		pre+=size[i];
		for(int j=0;j<=pre;j++)for(int k=0;k<2;k++)for(int o=0;o<2;o++)dp[j][k][o]=h[j][k][o];
	}
	for(int j=0;j<=size[x];j++)for(int k=0;k<2;k++)f[x][j][k]=0;
	for(int j=0;j<=pre;j++)for(int k=0;k<2;k++)for(int o=0;o<2;o++)
		up(f[x][j][k],dp[j][k][o]);
}
int main(){
	for(fac[0]=i=1;i<N;i++)fac[i]=1LL*fac[i-1]*i%P;
	for(inv[0]=inv[1]=1,i=2;i<N;i++)inv[i]=1LL*(P-inv[P%i])*(P/i)%P;
	for(i=2;i<N;i++)inv[i]=1LL*inv[i-1]*inv[i]%P;
	while(~scanf("%d%d",&n,&m)){
		if(!n)return 0;
		for(i=0;i<=n;i++)g[i]=0;
		for(i=2;i<=n;i++){
			scanf("%d",&x);
			nxt[i]=g[x];
			g[x]=i;
		}
		dfs(1);
		ans=0;
		for(i=0;i<=m;i++)up(ans,f[1][i][0]);
		printf("%d\n",ans);
	}
}
/*
5 1
1 1 5 1

3 2
1 1

0 0
*/

  

F. Fix and Solve

首先将每个数转化为质数出现最多一次的数。

对于每个质数用set维护所有出现的下标,那么相邻两个出现下标可以使得一段区间的非法。

线段树维护每个区间内被标记为非法的次数最小值以及最小值个数即可。

时间复杂度$O(n\log n\log\log n)$。

#include<cstdio>
#include<set>
#include<algorithm>
#include<vector>
using namespace std;
const int N=100010,M=262150;
int i,j,notprime[N];
vector<int>has[N];
set<int>T[N];
int n,K,m,a[N],all;
int tag[M],mi[M],mcnt[M];
void build(int x,int a,int b){
	mi[x]=tag[x]=0;
	mcnt[x]=b-a+1;
	if(a==b)return;
	int mid=(a+b)>>1;
	build(x<<1,a,mid);
	build(x<<1|1,mid+1,b);
}
inline void tag1(int x,int p){tag[x]+=p;mi[x]+=p;}
void change(int x,int a,int b,int c,int d,int p){
	if(c<=a&&b<=d){tag1(x,p);return;}
	if(tag[x]){
		tag1(x<<1,tag[x]);
		tag1(x<<1|1,tag[x]);
		tag[x]=0;
	}
	int mid=(a+b)>>1;
	if(c<=mid)change(x<<1,a,mid,c,d,p);
	if(d>mid)change(x<<1|1,mid+1,b,c,d,p);
	mi[x]=min(mi[x<<1],mi[x<<1|1]);
	mcnt[x]=0;
	if(mi[x]==mi[x<<1])mcnt[x]+=mcnt[x<<1];
	if(mi[x]==mi[x<<1|1])mcnt[x]+=mcnt[x<<1|1];
}
inline int query(){
	if(mi[1]>0)return all;
	return all-mcnt[1];
}
inline void gao(int l,int r,int p){
	if(r-l+1>K)return;
	int a=r-K+1,b=l;
	change(1,1,all,a,b,p);
}
inline void ins(int x,int y){
	T[x].insert(y);
	set<int>::iterator it=T[x].find(y),pre=it,nxt=it;
	pre--,nxt++;
	if(*pre&&*nxt<N)gao(*pre,*nxt,-1);
	if(*pre)gao(*pre,y,1);
	if(*nxt<N)gao(y,*nxt,1);
}
inline void del(int x,int y){
	set<int>::iterator it=T[x].find(y),pre=it,nxt=it;
	pre--,nxt++;
	if(*pre&&*nxt<N)gao(*pre,*nxt,1);
	if(*pre)gao(*pre,y,-1);
	if(*nxt<N)gao(y,*nxt,-1);
	T[x].erase(y);
}
inline void add(int x,int k){
	for(int i=0;i<has[k].size();i++)ins(has[k][i],x);
}
inline void remove(int x,int k){
	for(int i=0;i<has[k].size();i++)del(has[k][i],x);
}
int main(){
	for(i=2;i<N;i++){
		if(!notprime[i]){
			for(j=i;j<N;j+=i){
				notprime[j]=1;
				has[j].push_back(i);
			}
		}
	}
	while(~scanf("%d%d%d",&n,&K,&m)){
		if(!n)return 0;
		all=n-K+1;
		build(1,1,all);
		for(i=1;i<N;i++)T[i].clear(),T[i].insert(0),T[i].insert(N);
		for(i=1;i<=n;i++){
			scanf("%d",&a[i]);
			add(i,a[i]);
		}
		printf("%d\n",query());
		while(m--){
			int x,y;
			scanf("%d%d",&x,&y);
			remove(x,a[x]);
			add(x,a[x]=y);
			printf("%d\n",query());
		}
		long long ans=0;
		for(i=1;i<=n;i++)ans+=a[i];
		printf("%lld\n",ans);
	}
}

  

G. Gold Bandits

爆搜所有$1$到$2$的最短路,然后再求$1$到$2$的开销最小的最短路使得收益最大。

#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() {  }
#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 = 40, 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, m;
int d[N], f[N];
int v[N], w[N];
vector<int>a[N];
void bfs()
{
	queue<int>q; q.push(1); 
	MS(d, -1); d[1] = 0;
	while(!q.empty())
	{	
		int x = q.front(); q.pop();
		for(auto y : a[x])if(d[y] == -1)
		{
			q.push(y);
			d[y] = d[x] + 1;
		}
	}
}
bool e[N];
int ans;
int sub()
{
	MS(e, 0);
	MS(f, 63); f[2] = 0;
	for(int tim = 1; tim <= n; ++tim)
	{
		int x = 0;
		for(int i = 1; i <= n; ++i)if(!e[i] && f[i] < f[x])
		{
			x = i;
		}
		e[x] = 1;
		for(auto y : a[x])
		{
			gmin(f[y], f[x] + w[y]);
		}
	}
	return f[1];
}
void dfs(int x, int val)
{
	if(x == 2)
	{
		gmax(ans, val - sub());
		return;
	}
	for(auto y : a[x])if(d[y] == d[x] + 1)
	{
		w[y] = v[y];
		dfs(y, val + v[y]);
		w[y] = 0;
	}
}
int main()
{
	while(~scanf("%d%d", &n, &m), n || m)
	{
		for(int i = 1; i <= n; ++i)a[i].clear();
		for(int i = 3; i <= n; ++i)scanf("%d", &v[i]);
		for(int i = 1; i <= m; ++i)
		{
			int x, y; scanf("%d%d", &x, &y);
			a[x].push_back(y);
			a[y].push_back(x);
		}
		bfs();
		ans = 0;
		dfs(1, 0);
		printf("%d\n", ans);
	}
	return 0;
}
/*
【trick&&吐槽】
3 3
1
1 2
2 3
1 3

4 4
24 10
1 3
2 3
2 4
1 4

6 8
100 500 300 75
1 3
1 4
3 6
4 5
3 5
4 6
2 5
2 6

7 7
90 1000 700 2000 800
1 3
1 4
1 5
3 7
5 6
2 6
3 6

0 0

【题意】


【分析】


【时间复杂度&&优化】


*/

  

H. How Many Values?

$O(n\log a)$枚举所有$\gcd$本质不同的区间即可。

#include<cstdio>
const int N=100010;
int n,i,j,l[N],v[N],a[N],ans,vis[N];
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int main(){
	while(~scanf("%d",&n)){
		if(!n)return 0;
		for(i=0;i<=n;i++)a[i]=l[i]=v[i]=0;
		for(ans=i=0;i<=100;i++)vis[i]=0;
		for(i=1;i<=n;i++)scanf("%d",&a[i]);
		for(i=1;i<=n;i++)for(v[i]=a[i],j=l[i]=i;j;j=l[j]-1){
			v[j]=gcd(v[j],a[i]);
			while(l[j]>1&&gcd(a[i],v[l[j]-1])==gcd(a[i],v[j]))l[j]=l[l[j]-1];
			vis[v[j]]=1;
		}
		for(i=1;i<=100;i++)if(vis[i])ans++;
		printf("%d\n",ans);
	}
}

  

I. Integer Estate Agent

按题意模拟即可。

#include<cstdio>
typedef __int128 lll;
const int N = 1e6 + 10;
int n;
int cnt[N];

void init()
{
	int ans = 0;
	for(int i = 2; i <= 1e6; i ++){
		int x = 0;
		for(int j = 0; j <= 1e6; j ++){
			x += i + j;
			if(x > 1e6) break;
			cnt[x] ++;
			ans ++;
		}
	}
	//printf("%d\n", ans);
}
int main(){
	init();
	while(~ scanf("%d", &n), n){
		printf("%d\n", cnt[n]);
	}	
	return 0;
}

  

J. John and Super Mario 169

旅行商问题,状压DP即可,时间复杂度$O(2^nn^3)$。

#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() {  }
#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 = 1e5 + 10, M = 0, Z = 1e9 + 7;
const double inf = 1e18;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n;
struct P
{
	int x, y, z;
}me, sw[13], p[13][13];
double f[1 << 13][13];
double d[13][13];
double dp_sw[1 << 13][13];
double dp[1 << 13][13][13];
int K[13];
double sqr(double x)
{
	return x * x;
}
double dis(P a, P b)
{
	return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y) + sqr(a.z - b.z));
}
int main()
{
	while(~scanf("%d", &n), n)
	{
		scanf("%d%d%d", &me.x, &me.y, &me.z);
		for(int i = 0; i < n; ++i)
		{
			scanf("%d", &K[i]);
			scanf("%d%d%d", &sw[i].x, &sw[i].y, &sw[i].z);
			for(int j = 0; j < K[i]; ++j)
			{
				scanf("%d%d%d", &p[i][j].x, &p[i][j].y, &p[i][j].z);
			}
			
			int top = 1 << K[i];
			for(int j = 0; j < top; ++j)
			{
				for(int k = 0; k < K[i]; ++k)f[j][k] = inf;
			}
			for(int j = 0; j < K[i]; ++j)
			{
				f[1 << j][j] = dis(sw[i], p[i][j]);
			}
			for(int sta = 0; sta < top; ++sta)
			{
				for(int j = 0; j < K[i]; ++j)if(f[sta][j] != inf && (sta >> j & 1))
				{
					for(int k = 0; k < K[i]; ++k)if(~sta >> k & 1)
					{
						gmin(f[sta | 1 << k][k], f[sta][j] + dis(p[i][j], p[i][k]));
					}
				}
			}
			
			for(int j = 0; j < K[i]; ++j)
			{
				d[i][j] = f[top - 1][j]; 
			}
		}
		
		int top = 1 << n;
		for(int sta = 0; sta < top; ++sta)
		{
			for(int j = 0; j < n; ++j)
			{
				dp_sw[sta][j] = inf;
				for(int k = 0; k < K[j]; ++k)
				{
					dp[sta][j][k] = inf;
				}
			}
		}
		for(int j = 0; j < n; ++j)
		{
			dp_sw[1 << j][j] = dis(me, sw[j]);
		}
		for(int sta = 0; sta < top; ++sta)
		{
			for(int j = 0; j < n; ++j)
			{
				if(dp_sw[sta][j] != inf)
				{
					for(int k = 0; k < K[j]; ++k)
					{
						gmin(dp[sta][j][k], dp_sw[sta][j] + d[j][k]);
					}
				}
			}
			for(int j = 0; j < n; ++j)if(sta >> j & 1)
			{
				for(int u = 0; u < n; ++u)if(~sta >> u & 1)
				{
					for(int k = 0; k < K[j]; ++k)if(dp[sta][j][k] != inf)
					{
						gmin(dp_sw[sta | 1 << u][u], dp[sta][j][k] + dis(p[j][k], sw[u]));
						/*
						for(int v = 0; v < K[u]; ++v)
						{
							gmin(dp[sta | 1 << u][u][v], 
							dp[sta][j][k] + dis(p[j][k], sw[u]) + d[u][v]);
						}
						*/
					}
				}
			}
		}
		double ans = inf;
		for(int j = 0; j < n; ++j)
		{	
			for(int k = 0; k < K[j]; ++k)
			{
				gmin(ans, dp[top - 1][j][k]);
			}
		}
		printf("%.10f\n", ans);
	}
	return 0;
}
/*
【trick&&吐槽】
2 5 5 0
4 6 0 0
7 0 0
-11 -1 0
-11 1 0

-10 0 0
2 5 0 0
0 0 0
0 5 0

0 0 0 0

【题意】


【分析】


【时间复杂度&&优化】


*/

  

posted @ 2018-04-14 02:04 Claris 阅读(...) 评论(...) 编辑 收藏