BestCoder Round#77

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今天开始将要转一下做题的OJ了。。
虽然感觉放着权限号在那里不好。。。。。

然后先写一下BC #77的题解吧

1001
就是在枚举子集的过程中我们发现当n > 1时，x出现了偶数次，所以其对答案的贡献就是0；当 n = 1时，其对答案的贡献是 x

1002
简单的排列组合

1003
考虑离线之后我们就可以用并查集做了

1004
DP 是可以用单调队列优化成O(n)的不过我打的是O(n2)的

1005
这一题可以考虑离线
离线做法很经典先将询问按R排序然后就可以在树状数组上修改了
在线的话代码量可能比较大就是用可持久话线段树具体思想和那个询问[l,r]区间的边形成几个联通块差不多

总的来说这一场还是比较水的
PS：感谢这一场带我飞的Manchery大爷

1001

#include<cstdio>#include<cstdlib>using namespace std;inline char nc(){    static char buf[100000],*p1=buf,*p2=buf;    if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }    return *p1++;}#define ll long longinline void read(ll &x){    char c=nc(),b=1;    for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;    for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}ll n,a,ans;int main(){    ll Q;    read(Q);    while (Q--)    {        read(n);        for (ll i=1;i<=n;i++)            read(a);        if (n>1)             ans=0;        else             ans=a;        printf("%I64d\n",ans);    }    return 0;}

1002

#include <cstdio>#include <cstdlib>#include <iostream>#include <algorithm>#include <cstring>const int maxn = 1005;const int mod = 1e9 + 7;int n;int cnt[30];char s[maxn];int c[maxn][maxn];int main(){    int T;    scanf("%d", &T);    c[1][0] = c[1][1] = 1;    for(int i = 2; i <= 1000; i++)    {        c[i][0] = 1;        for(int j = 1; j <= i; j++)            (c[i][j] = c[i - 1][j - 1] + c[i - 1][j]) %= mod;    }    while(T--)    {        scanf("%s", s + 1);        n = strlen(s + 1);        memset(cnt, 0, sizeof(cnt));        for(int i = 1; i <= n; i++)    cnt[s[i] - 'a' + 1]++;        int odd = 0;        for(int i = 1; i <= 26; i++) odd += cnt[i] & 1;        if(odd > 1) puts("0");        else        {            int tot = 0;            long long ans = 1;            for(int i = 1; i <= 26; i++)            {                    cnt[i] /= 2;                if(cnt[i])                {                    (ans *= c[cnt[i] + tot][cnt[i]]) %= mod;                    tot += cnt[i];                }            }            cout << ans << std;        }    }    return 0;}

1003

#include<cstdio>#include<cstdlib>#include<cstring>#define cl(x) memset(x,0,sizeof(x))using namespace std;#define ll long longinline char nc(){    static char buf[100000],*p1=buf,*p2=buf;    if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }    return *p1++;}inline void read(int &x){    char c=nc(),b=1;    for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;    for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}inline int read(char *s){    char c=nc(); int len=0;    for (;!(c>='0' && c<='1');c=nc());    for (;c>='0' && c<='1';s[++len]=c,c=nc()); s[++len]=0; return len-1;}int n,m,S,T,K;int fat[250005];char M[505][505];int X[250005],Y[250005];const int dx[]={0,0,1,-1};const int dy[]={1,-1,0,0};inline int P(int x,int y){    if (x==0) return S;    if (x==n+1) return T;    return (x-1)*m+y;}inline int getfat(int u){    return u==fat[u]?u:fat[u]=getfat(fat[u]);}inline void link(int x,int y){    int fx=getfat(x),fy=getfat(y);    if (fx!=fy)        fat[fx]=fy;}inline bool Jud(int x,int y){    int fx=getfat(x),fy=getfat(y);    return fx==fy;}int main(){    int Q;    read(Q);    while (Q--)    {        read(n); read(m); S=n*m+1; T=n*m+2;        for (int i=1;i<=n;i++)            read(M[i]);        read(K);        for (int i=1;i<=K;i++) read(X[i]),read(Y[i]),X[i]++,Y[i]++,M[X[i]][Y[i]]='1';        for (int i=1;i<=T;i++) fat[i]=i;        for (int i=1;i<=n;i++)            for (int j=1;j<=m;j++)                if (M[i][j]!='1')                {                    int a,b;                    for (int k=0;k<4;k++)                    {                        a=i+dx[k],b=j+dy[k];                        if (b>0 && b<=m)                            if (a==0 || a==n+1 || M[a][b]!='1')                                link(P(a,b),P(i,j));                    }                }        if (Jud(S,T)){            printf("-1\n");            continue;        }        int flag=0;        for (int i=K;i && !flag;i--)        {            int a,b;            for (int k=0;k<4;k++)            {                a=X[i]+dx[k],b=Y[i]+dy[k];                if (b>0 && b<=m)                    if (a==0 || a==n+1 || M[a][b]!='1')                        link(P(a,b),P(X[i],Y[i]));            }            M[X[i]][Y[i]]='0';            if (Jud(S,T)){                printf("%d\n",i);                flag=1;            }        }        if (!flag)            printf("0\n");        }    return 0;}

1004

#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#define cl(x) memset(x,0,sizeof(x))using namespace std;#define ll long longinline char nc(){    static char buf[100000],*p1=buf,*p2=buf;    if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }    return *p1++;}inline void read(int &x){    char c=nc(),b=1;    for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;    for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}int n,m;int sum[2005];double f[2005];inline double Log(int x){    return log((double)x)/log(2.0);}int main(){    int Q,a;    read(Q);    while (Q--)    {        cl(sum); cl(f);        read(m); read(n);        for (int i=1;i<=n;i++) read(a),sum[++a]=1;        for (int i=1;i<=m;i++) sum[i]+=sum[i-1];        for (int i=1;i<=m;i++)            for (int j=i-1;j>=0;j--)            {                if (sum[i]-sum[j]>1) break;                if (sum[i]-sum[j]<1) continue;                f[i]=max(f[i],f[j]+Log(i-j));            }        printf("%I64d\n",(ll)(f[m]*1000000.0));    }    return 0;}

1005
在线打法

#include<cstdio>#include<iostream>#include<cstring>#include<cstdlib>using namespace std;struct Rec{   int a,b,c;   inline friend bool operator !=(Rec a,Rec b)   {return (a.a^b.a)|(a.b^b.b)|(a.c^b.c);}};const   int MaxHash=213133;inline int Calc(Rec Key){    return ((Key.a*3+Key.b*76721+Key.c*23)%MaxHash+MaxHash)%MaxHash;}struct Hash{   Rec Key;   int Latest;   Hash*next;}*Table[MaxHash];Hash Hash_Cache[MaxHash];int Hash_Tot;inline Hash*New_Hash(){return Hash_Cache+Hash_Tot++;}Hash *Ans;bool find(Rec Key,Hash*Cur){    if(!Cur)return false;    if(Cur->Key!=Key)return find(Key,Cur->next);    Ans=Cur;return true;}inline void insert(Rec Key,int place,int time){   Hash*tp=New_Hash();   tp->next=Table[place];   Table[place]=tp;   tp->Latest=time;   tp->Key=Key;}struct Seg{    Seg *lc,*rc;    int l,r;    int data;}*Root[200001];Seg Seg_Cache[5000000];int Seg_Tot;inline Seg*New_Seg(){return Seg_Cache+Seg_Tot++;}int Query(Seg*Cur,int L){   if(!Cur)return 0;   if(Cur->r<L)return 0;   else if(Cur->l>=L)return Cur->data;    else      {        int res=0;        res=Query(Cur->lc,L);        res+=Query(Cur->rc,L);        return res;     }}Seg*Modify(Seg*Cur,int L){    Seg*Ne=New_Seg();     *Ne=*Cur;    Ne->data++;     if(Cur->l^Cur->r)     {      int Mid=Cur->l+Cur->r>>1;      if(Mid<L)        Ne->rc=Modify(Ne->rc,L);      else        Ne->lc=Modify(Ne->lc,L);     }   return Ne;}Seg*Build(int L,int R){    Seg*Ne=New_Seg();     Ne->l=L;Ne->r=R;    Ne->lc=Ne->rc=NULL;    Ne->data=0;     if(L^R)     {        int Mid=L+R>>1;        Ne->rc=Build(Mid+1,R);        Ne->lc=Build(L,Mid);     }   return Ne;}int n,Q;int A[200001];char c;inline void read(int &a){  a=0;do c=getchar();while(c<'0'||c>'9');  while(c<='9'&&c>='0')a=(a<<3)+(a<<1)+c-'0',c=getchar();}int Val[200001];int main(){    Rec tp;     int pl,T;    read(T);    while(T--)  {      Hash_Tot=Seg_Tot=0;    memset(Table,0,sizeof(Table));    read(n);     Root[1]=Root[0]=Root[2]=Build(1,n);    for(int i=1;i<=n;i++)     read(A[i]);    for(int i=3;i<=n;i++)       {         tp.a=A[i-2],tp.b=A[i-1],tp.c=A[i];          Val[i]=0;        if(tp.a<=tp.b&&tp.b<=tp.c)         if(!find(tp,Table[pl=Calc(tp)]))              insert(tp,pl,i),Root[i]=Modify(Root[i-1],1);         else           {              Root[i]=Modify(Root[i-1],Ans->Latest);              Ans->Latest=i;                 }        else Val[i]=-1,Root[i]=Root[i-1];          Val[i]=Val[i-1]+Val[i]+1;      }   read(Q);    int L,R;    while(Q--)    {       read(L),read(R);       if(R-L<=1)           {puts("0");continue;}       int  ans=Query(Root[R],L+2);          ans=Val[R]-Val[L+1]-ans;         printf("%d\n",ans);   } }    getchar();    getchar();    getchar();    getchar();    getchar();    getchar();    getchar();    getchar();    return 0;}

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