两次考试

11.6

T1

super_gcd

还是要抓紧复习一下板子...

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define rint register int
using namespace std;
inline void read(int &x)
{
    x=0; int ff=1; char q=getchar();
    while(q<'0'||q>'9') { if(q=='-') ff=-1; q=getchar(); }
    while(q>='0'&&q<='9') x=x*10+q-'0',q=getchar();
    x*=ff;
}
const int LEN=1006;

int t[LEN],he;
int cm;

struct bign
{
    int s[LEN],len;
    bign(){mem(s,0);len=1;}
    void read()
    {
        int i;
        he=0; char q=getchar();
        while(q<'0'||q>'9') q=getchar();
        while(q>='0'&&q<='9') t[++he]=q-'0',q=getchar();
        for(i=he;i;--i) s[he-i+1]=t[i];
        len=he;
    }
    bool operator < (const bign &c) const
    {
        bign x=*this;
        if(x.len==c.len)
        {
            int i;
            for(i=x.len;i;--i)
            {
                if(x.s[i]==c.s[i]) continue;
                return x.s[i]<c.s[i];
            }
        }
        else
            return x.len<c.len;
        return 0;
    }
    bign operator - (const bign &c) const
    {
        bign x=*this; int i;
        for(i=1;i<=c.len;++i)
        {
            x.s[i+cm]-=c.s[i];
            if(x.s[i+cm]<0)
            {
                --x.s[i+cm+1];
                x.s[i+cm]+=10;
            }
        }
        while(x.len>1&&x.s[x.len]==0)
            --x.len;
        return x;
    }
    void out()
    {
        int i;
        for(i=len;i;--i)
            printf("%d",s[i]);
        puts("");
    }
};

int T;
bign A,B;

int gcd()
{
    while( !(A.len==1&&A.s[1]==0)&&!(B.len==1&&B.s[1]==0) )
    {
        if(A<B)
        {
            cm=B.len-A.len-1;
            if(cm<0) cm=0;
            B=B-A;
        }
        else
        {
            cm=A.len-B.len-1;
            if(cm<0) cm=0;
            A=A-B;
        }
    }
    //A.out(); B.out();
    if(A<B)
    {
        if(B.len==1&&B.s[1]==1&&A.len==1&&A.s[1]==0)
            return 1;
        return 0;
    }
    else
    {
        if(A.len==1&&A.s[1]==1&&B.len==1&&B.s[1]==0)
            return 1;
        return 0;
    }
}

int main(){
    
    //freopen("T1.in","r",stdin);
    
    //freopen("king.in","r",stdin);
    //freopen("king.out","w",stdout);
    
    read(T);
    while(T--)
    {
        A.read(); B.read();
        //A.out(); B.out();
        if(gcd())
            puts("Yes");
        else
            puts("No");
    }
}
T1

T2

考试什么都想不出来...

考虑补集转化,用总方案数减去0、1、2各自不合法的方案数

因为每个区间只会有一个数超过一半,所以不用容斥

枚举0、1、2

把跟它相等的数看做1,不相等的看做-1

这样统计前缀和中比当前低的位置即可

$O(n)$

其实$O(nlogn)$的很好想

$$pre_r-pre_{l-1}>\frac{r-l+1}{2}$$

$$2*pre_r-r>2*pre_{l-1}-l+1$$

直接树状数组就行了

真是菜...

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define rint register int
using namespace std;
inline void read(int &x)
{
    char q=getchar();
    while(q<'0'||q>'9') q=getchar();
    x=q-'0';
}
const int N=5000006;

int n;
int cm;
int aee=5000000;
int cnt[N<<1],v[N];
ll ans;

ll get()
{
    rint i; int tt; ll ans=0,now=aee,ccc=0;
    mem(cnt,0); ++cnt[aee];
    for(i=1;i<=n;++i)
    {
        if(v[i]==cm) ccc+=cnt[now],++now;
        else --now,ccc-=cnt[now];
        ++cnt[now];
        ans+=ccc;
    }
    return ans;
}

int main(){
    
    //freopen("ex1.in","r",stdin);
    
    rint i;
    
    scanf("%d",&n);
    for(i=1;i<=n;++i) read(v[i]);
    
    /*printf("\n");
    for(i=1;i<=n;++i)
        printf("%d",v[i]);
    printf("\n");*/
    
    ans=1LL*n*(n+1)/2;
    for(i=0;i<3;++i)
        cm=i,ans-=get();
    printf("%lld\n",ans);
}
T2

T3

$O(n^3)$的dp很简单,但是我根本没想...

考试打了用堆来贪心取,骗了40

70分 还可以打网络流

建图就S向豆干和干脆面连 流量为给出个数费用为0的边

豆干和干脆面在向每只猫连 一条$1/p_i$ 一条$1/0$ 的边

正解是wqs二分套wqs二分

二分两个cost,分别表示选豆干和干脆面各会需要额外的花费

每次dp记录最优的猫的个数和用掉的豆干和干脆面的个数

这样一直二分到第一个大于等于给出个数的值

而且二分出的值不一定可以使得dp结果正好是给出的个数

就像陈立杰那个黑白树一样(毕竟重打了7遍呢,记忆比较深刻...)

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define eps 0.00000001
#define dd double
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define rint register int
using namespace std;
inline void read(int &x)
{
    x=0; int ff=1; char q=getchar();
    while(q<'0'||q>'9') { if(q=='-') ff=-1; q=getchar(); }
    while(q>='0'&&q<='9') x=x*10+q-'0',q=getchar();
    x*=ff;
}
const int N=100006;
const ll Inf=1e10;

int n,A,B;
int now,pr;
dd p[N],q[N];
dd f[2];
int ga[2],gb[2];

void check(dd xp,dd xq)
{
    rint i,j;
    now=0;
    f[0]=0; ga[0]=gb[0]=0;
    for(i=1;i<=n;++i)
    {
        now^=1; pr=now^1;
        f[now]=-Inf; ga[now]=gb[now]=0;
        if(f[now]<f[pr])
            f[now]=f[pr],ga[now]=ga[pr],gb[now]=gb[pr];
        if(f[now]<f[pr]+p[i]-xp)
            f[now]=f[pr]+p[i]-xp,ga[now]=ga[pr]+1,gb[now]=gb[pr];
        if(f[now]<f[pr]+q[i]-xq)
            f[now]=f[pr]+q[i]-xq,ga[now]=ga[pr],gb[now]=gb[pr]+1;
        if(f[now]<f[pr]+p[i]+q[i]-p[i]*q[i]-xp-xq)
            f[now]=f[pr]+p[i]+q[i]-p[i]*q[i]-xp-xq,ga[now]=ga[pr]+1,gb[now]=gb[pr]+1;
    }
}

dd work()
{
    dd l=0,r=1,mid,l2,r2,mid2;
    while(l<r-eps)
    {
        mid=(l+r)/2.0;
        //printf("l=%f r=%f mid=%f\n",l,r,mid);
        l2=0; r2=1;
        while(l2<r2-eps)
        {
            mid2=(l2+r2)/2.0;
            //printf("l2=%f r2=%f mid2=%f gb[now]=%d\n",l2,r2,mid2,gb[now]);
            check(mid,mid2);
            if(gb[now]==B)
                break;
            if(gb[now]<B) r2=mid2;
            else l2=mid2;
        }
        if(ga[now]==A)
            break;
        if(ga[now]<A) r=mid;
        else l=mid;
    }
    return f[now]+mid*A+mid2*B;
}

int main(){
    
    //freopen("in.in","r",stdin);
    
    //freopen("red8.in","r",stdin);
    
    rint i;
    
    while(~scanf("%d%d%d",&n,&A,&B))
    {
        for(i=1;i<=n;++i) scanf("%lf",&p[i]);
        for(i=1;i<=n;++i) scanf("%lf",&q[i]);
        if(A>n) A=n;
        if(B>n) B=n;
        printf("%.3f\n",work());
    }
}
T3

11.7

T1

T一定可以分解成$S*B^{a}+n1*A*B^{a-1}+n2*A*B^{a-2}...$的形式

其中n1,n2是系数,可以为0

这样的话,直接枚举a,在贪心取就行了

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define rint register int
using namespace std;
inline void read(int &x)
{
    x=0; int ff=1; char q=getchar();
    while(q<'0'||q>'9') { if(q=='-') ff=-1; q=getchar(); }
    while(q>='0'&&q<='9') x=x*10+q-'0',q=getchar();
    x*=ff;
}

int S,T,A,B;
int maxk;
ll P[106];

int main(){
    
    //freopen("T1.in","r",stdin);
    //freopen("T1hh.out","w",stdout);
    
    //freopen("a.in","r",stdin);
    //freopen("a.out","w",stdout);
    
    rint i,j; ll tt,t1; int con,ans=1e9;
    
    read(S); read(T); read(A); read(B);
    
    if(S>T)
    {
        puts("-1");
        return 0;
    }
    
    tt=S; maxk=0;
    while(tt<=T&&maxk<=32) tt*=B,++maxk;
    --maxk; tt/=B;
    
    P[0]=1;
    for(i=1;i<=maxk;++i)
        P[i]=P[i-1]*B;
    
    for(i=maxk;~i;--i)
    {
        tt=S*P[i]; con=i;
        for(j=i;~j;--j)
        {
            t1=(1LL*T-tt)/(1LL*A*P[j]);
            con+=t1; tt+=1LL*t1*A*P[j];
        }
        if(tt==T&&ans>con) ans=con;
    }
    if(ans==1000000000)
        ans=-1;
    printf("%d\n",ans);
}
T1

T2

算法1(考试骗分) 90

先考虑二分ans

然后二分点对的左端点距离选择的边的左端点的最大值

然后对右端点做区间交

$O(nlog^2n)$

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define rint register int
using namespace std;
inline void read(int &x)
{
    x=0; int ff=1; char q=getchar();
    while(q<'0'||q>'9') { if(q=='-') ff=-1; q=getchar(); }
    while(q>='0'&&q<='9') x=x*10+q-'0',q=getchar();
    x*=ff;
}
const int N=100006;

int n,m;
int u[N],v[N],len[N];

int check60(int maxd)
{
    rint i,j; int mn,mx,t1,t2;
    for(i=1;i<=n;++i)
    {
        mn=1000000; mx=1;
        for(j=1;j<=m;++j)
            if(len[j]>maxd)
            {
                t1=u[j]-i; if(t1<0) t1=-t1;
                if(t1>maxd)
                {
                    mn=1; mx=1000000;
                    break;
                }
                t2=maxd-t1;
                if(mn>v[j]+t2) mn=v[j]+t2;
                if(mx<v[j]-t2) mx=v[j]-t2;
            }
        if(mx<=mn) return 1;
    }
    return 0;
}

int work60()
{
    int l=0,r=n,mid,ans=n;
    while(l<=r)
    {
        mid=(l+r)>>1;
        //printf("l=%d r=%d mid=%d\n",l,r,mid);
        if(check60(mid)) ans=mid,r=mid-1;
        else l=mid+1;
    }
    return ans;
}

int check100(int maxd)
{
    int l=0,r=maxd,mid;
    int mn,mx,t1,t2,tt,l1=n,r1=1;
    rint i,j;
    while(l<=r)
    {
        mid=(l+r)>>1;
        //printf("l2=%d r2=%d mid2=%d\n",l,r,mid);
        mn=n; mx=1;
        for(i=1;i<=m;++i)
            if(len[i]>maxd)
            {
                if(mx<u[i]-mid) mx=u[i]-mid;
                if(mn>u[i]+mid) mn=u[i]+mid;
            }
        //printf("mn=%d mx=%d\n",mn,mx);
        if(mx>mn) l=mid+1;
        else
        {
            l1=mx,r1=mn,r=mid-1;
            /*if(l1>mx) l1=mx;
            if(r1<mn) r1=mn;
            r=mid-1;*/
            /*if(r1-l1<mn-mx)
                l1=mx,r1=mn;
            r=mid-1;*/
        }
    }
    if(l1>r1) return 0;
    //printf("l1=%d r1=%d mid=%d\n",l1,r1,mid);
    for(i=l1;i<=r1;++i)
    {
        t1=n; t2=1;
        for(j=1;j<=m;++j)
            if(len[j]>maxd)
            {
                tt=u[j]-i; if(tt<0) tt=-tt;
                tt=maxd-tt;
                if(t2<v[j]-tt) t2=v[j]-tt;
                if(t1>v[j]+tt) t1=v[j]+tt;
            }
        //printf("asdsd:: %d %d\n",t2,t1);
        if(t2<=t1) return 1;
    }
    return 0;
}

int work100()
{
    int l=0,r=n,mid,ans=n;
    while(l<=r)
    {
        mid=(l+r)>>1;
        //printf("l=%d r=%d mid=%d\n",l,r,mid);
        if(check100(mid)) ans=mid,r=mid-1;
        else l=mid+1;
    }
    return ans;
}

int main(){
    
    //freopen("T2.in","r",stdin);
    //freopen("T2.out","w",stdout);
    
    //freopen("b.in","r",stdin);
    //freopen("b.out","w",stdout);
    
    rint i;
    
    read(n); read(m);
    for(i=1;i<=m;++i)
        read(u[i]),read(v[i]),len[i]=v[i]-u[i];
    if(n<=5000)
        printf("%d\n",work60());
    else
        printf("%d\n",work100());
}
1

算法2 100

其实左端点并不具有二分性质,而是一个横坐标为左端点pos,竖坐标为ans的下凹函数

那么可以三分套二分

$O(nlog^2n)$

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define rint register int
using namespace std;
inline void read(int &x)
{
    x=0; int ff=1; char q=getchar();
    while(q<'0'||q>'9') { if(q=='-') ff=-1; q=getchar(); }
    while(q>='0'&&q<='9') x=x*10+q-'0',q=getchar();
    x*=ff;
}
const int N=100006;

int n,m;
int u[N],v[N],len[N];

int check2(int pos,int maxd)
{
    rint i; int l1=1,r1=n,t1;
    for(i=1;i<=m;++i)
        if(len[i]>maxd)
        {
            t1=u[i]-pos; if(t1<0) t1=-t1;
            if(t1>maxd) return 0;
        }
    for(i=1;i<=m;++i)
        if(len[i]>maxd)
        {
            t1=u[i]-pos; if(t1<0) t1=-t1;
            t1=maxd-t1;
            if(l1<v[i]-t1) l1=v[i]-t1;
            if(r1>v[i]+t1) r1=v[i]+t1;
        }
    if(l1<=r1) return 1;
    return 0;
}

int check(int pos)
{
    rint i;
    int l=0,r=n,mid,ans=n;
    while(l<=r)
    {
        mid=(l+r)>>1;
        if(check2(pos,mid)) ans=mid,r=mid-1;
        else l=mid+1;
    }
    return ans;
}

int work()
{
    int l=1,r=n,mid,midmid,midv,midmidv,ans=n,i;
    while(l<=r-10)
    {
        mid=l+(r-l)/3; midmid=r-(r-l)/3;
        midv=check(mid); midmidv=check(midmid);
        if(midv>midmidv)
        {
            if(ans>midmidv) ans=midmidv;
            l=mid;
        }
        else
        {
            if(ans>midv) ans=midv;
            r=midmid;
        }
    }
    for(i=l;i<=r;++i)
    {
        midv=check(i);
        if(ans>midv) ans=midv;
    }
    return ans;
}

int main(){
    
    //freopen("T2.in","r",stdin);
    
    rint i;
    
    read(n); read(m);
    for(i=1;i<=m;++i)
        read(u[i]),read(v[i]),len[i]=v[i]-u[i];
    printf("%d\n",work());
}
2

算法3 100

二分ans

对于每一个点对,发现合法的区间映射到二维平面是其实是一个个倾斜45度的正方形

用$(x+y,x-y)$把正方形旋转成水平的(旋转过程大小会变,但是相对位置不会)

这样矩形求交就行了

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define rint register int
using namespace std;
inline void read(int &x)
{
    x=0; int ff=1; char q=getchar();
    while(q<'0'||q>'9') { if(q=='-') ff=-1; q=getchar(); }
    while(q>='0'&&q<='9') x=x*10+q-'0',q=getchar();
    x*=ff;
}
const int N=100006;
const int Inf=1e9;

int n,m;
int u[N],v[N],len[N];

int check(int maxd)
{
    rint i;
    int l=-Inf,r=Inf,xi=-Inf,sh=Inf;
    int t1,t2;
    int sha,xia,zuo,you;
    for(i=1;i<=m;++i)
        if(len[i]>maxd)
        {
            sha=-Inf; xia=Inf; zuo=Inf; you=-Inf;
            t1=u[i]-maxd+v[i]; t2=u[i]-maxd-v[i];
            if(sha<t2) sha=t2; if(xia>t2) xia=t2;
            if(you<t1) you=t1; if(zuo>t1) zuo=t1;
            t1=u[i]+maxd+v[i]; t2=u[i]+maxd-v[i];
            if(sha<t2) sha=t2; if(xia>t2) xia=t2;
            if(you<t1) you=t1; if(zuo>t1) zuo=t1;
            t1=u[i]+v[i]-maxd; t2=u[i]-(v[i]-maxd);
            if(sha<t2) sha=t2; if(xia>t2) xia=t2;
            if(you<t1) you=t1; if(zuo>t1) zuo=t1;
            t1=u[i]+v[i]+maxd; t2=u[i]-(v[i]+maxd);
            if(sha<t2) sha=t2; if(xia>t2) xia=t2;
            if(you<t1) you=t1; if(zuo>t1) zuo=t1;
            
            if(l<zuo) l=zuo; if(r>you) r=you;
            if(xi<xia) xi=xia; if(sh>sha) sh=sha;
        }
    if(xi<=sh&&l<=r)
        return 1;
    return 0;
}

int work()
{
    int l=0,r=n,mid,ans=n;
    while(l<=r)
    {
        mid=(l+r)>>1;
        //printf("l=%d r=%d mid=%d\n",l,r,mid);
        if(check(mid)) ans=mid,r=mid-1;
        else l=mid+1;
    }
    return ans;
}

int main(){
    
    //freopen("T2.in","r",stdin);
    //freopen("T2.out","w",stdout);
    
    rint i;
    
    read(n); read(m);
    for(i=1;i<=m;++i)
        read(u[i]),read(v[i]),len[i]=v[i]-u[i];
    printf("%d\n",work());
}
3

算法4 100

其实算法3的check就是求曼哈顿距离可否满足maxd

那么把曼哈顿距离转化成切比雪夫距离

这样横纵坐标就独立起来了

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define rint register int
using namespace std;
inline void read(int &x)
{
    x=0; int ff=1; char q=getchar();
    while(q<'0'||q>'9') { if(q=='-') ff=-1; q=getchar(); }
    while(q>='0'&&q<='9') x=x*10+q-'0',q=getchar();
    x*=ff;
}
const int Inf=1e9;
const int N=100006;

int n,m;
int u[N],v[N],len[N];

int check(int maxd)
{
    rint i; int l=-Inf,r=Inf;
    for(i=1;i<=m;++i)
        if(len[i]>maxd)
        {
            if(l<u[i]+v[i]-maxd) l=u[i]+v[i]-maxd;
            if(r>u[i]+v[i]+maxd) r=u[i]+v[i]+maxd;
        }
    //printf("l1=%d r1=%d\n",l,r);
    if(l>r) return 0;
    l=-Inf; r=Inf;
    for(i=1;i<=m;++i)
        if(len[i]>maxd)
        {
            if(l<u[i]-v[i]-maxd) l=u[i]-v[i]-maxd;
            if(r>u[i]-v[i]+maxd) r=u[i]-v[i]+maxd;
        }
    //printf("l2=%d r2=%d\n",l,r);
    if(l>r) return 0;
    return 1;
}

int work()
{
    int l=0,r=n,mid,ans=n;
    while(l<=r)
    {
        mid=(l+r)>>1;
        //printf("l=%d r=%d mid=%d\n",l,r,mid);
        if(check(mid)) ans=mid,r=mid-1;
        else l=mid+1;
    }
    return ans;
}

int main(){
    
//    freopen("T2.in","r",stdin);
    
    rint i;
    read(n); read(m);
    for(i=1;i<=m;++i)
        read(u[i]),read(v[i]),len[i]=v[i]-u[i];
    printf("%d\n",work());
}
4

T3

结论:

不管对手怎么动,只要预先知道了他的行动

就可以对应的进行行动,总有一种方案使得最后得到结果

二分即可

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define rint register int
using namespace std;
inline void read(int &x)
{
    x=0; int ff=1; char q=getchar();
    while(q<'0'||q>'9') { if(q=='-') ff=-1; q=getchar(); }
    while(q>='0'&&q<='9') x=x*10+q-'0',q=getchar();
    x*=ff;
}
const int N=300006;
int first[N],nt[N<<1],ver[N<<1],e;
void addb(int u,int v)
{
    ver[e]=v;
    nt[e]=first[u];
    first[u]=e++;
}

int n,alln;
int v[N],t[N],pos[N];
int x[N<<1],y[N<<1];

int dfn[N],low[N],tim,zhan[N],he,sun[N],sum;
bool flag[N];
void tarjan(int x)
{
    dfn[x]=low[x]=++tim;
    zhan[++he]=x; flag[x]=1;
    int i;
    for(i=first[x];i!=-1;i=nt[i])
    {
        if(dfn[ver[i]]==-1)
        {
            tarjan(ver[i]);
            if(low[x]>low[ver[i]])
                low[x]=low[ver[i]];
        }
        else
            if(flag[ver[i]]&&low[x]>dfn[ver[i]])
                low[x]=dfn[ver[i]];
    }
    if(dfn[x]==low[x])
    {
        i=-1; ++sum;
        while(i!=x)
            i=zhan[he--],flag[i]=0,++sun[sum];
    }
}

int check(int arr)
{
    rint i; int tt;
    for(i=0;i<n;++i) t[i]=v[i];
    for(i=1;i<=arr;++i) swap(t[x[i]],t[y[i]]);
    
    /*printf("\n");
    for(i=0;i<n;++i)
        printf("%d ",t[i]);
    printf("\n");*/
    
    for(i=0;i<n;++i) pos[t[i]]=i;
    for(i=0;i<n;++i) sun[i]=0,dfn[i]=-1,first[i]=-1;
    e=0; tim=0; he=0; sum=0;
    for(i=0;i<n;++i) addb(i,pos[i]);
    for(i=0;i<n;++i)
        if(dfn[i]==-1)
            tarjan(i);
    tt=0;
    
    /*printf("\n");
    for(i=1;i<=sum;++i)
        printf("%d ",sun[i]);
    printf("\n");*/
    
    for(i=1;i<=sum;++i)
        tt+=(sun[i]-1);
    //printf("arr=%d tt=%d\n",arr,tt);
    return tt<=arr;
}

int work()
{
    int l=0,r=alln,mid,ans=alln;
    while(l<=r)
    {
        mid=(l+r)>>1;
        if(check(mid)) ans=mid,r=mid-1;
        else l=mid+1;
    }
    return ans;
}

int main(){
    
    //freopen("T3.in","r",stdin);
    
    rint i;
    
    read(n); alln=n<<1;
    for(i=0;i<n;++i)
        read(v[i]);
    for(i=1;i<=alln;++i)
        read(x[i]),read(y[i]);
    printf("%d\n",work());
}
T3

 

posted @ 2017-11-08 07:00  A_LEAF  阅读(200)  评论(0编辑  收藏  举报