首页 > 代码库 > 【CF刷题】14-05-12

【CF刷题】14-05-12

  Round 236 div.1

    A:只需要每个点连接所有比他大的点,知道边用完为止。  

//By BLADEVIL
#include <cmath>
#include <cstdio>
#define maxn 25;

using namespace std;


int main() {
    int task; scanf("%d",&task);
    while (task--) {
        int n,p; scanf("%d%d",&n,&p);
        int rest(2*n+p);
        for (int i=1;(i<=n)&&rest;i++)
            for (int j=i+1;(j<=n)&&rest;j++) printf("%d %d\n",i,j),rest--;
    } 
    return 0;
}
View Code

    B:求出g[i]=gcd(a[1],a[2]....a[i]),那么只需要从后向前贪心的考虑每个g[i]能不能被除掉就好了。

//By BLADEVIL
#include <cstdio>
#include <algorithm>
#define maxn 5010

using namespace std;

int n,m;
int a[maxn],b[maxn],g[maxn];

int calc(int x) {
    int ans(0);
    for (int i=1;i<=m;i++) while (!(x%b[i])) ans--,x/=b[i];
    for (int i=2;i*i<=x;i++) while (!(x%i)) ans++,x/=i;
    if (x>1) ans++;
    return ans;
}

int main() {
    scanf("%d%d",&n,&m);
    for (int i=1;i<=n;i++) scanf("%d",&a[i]),g[i]=__gcd(g[i-1],a[i]);
    for (int i=1;i<=m;i++) scanf("%d",&b[i]);
    int cur(1);
    for (int i=n;i;i--) {
        a[i]/=cur; g[i]/=cur;
        if (calc(g[i])<0) cur*=g[i],a[i]/=g[i];
    }
    //for (int i=1;i<=n;i++) printf("%d ",a[i]); printf("\n");
    int ans(0);
    for (int i=1;i<=n;i++) ans+=calc(a[i]);
    printf("%d\n",ans);
    return 0;
}
View Code

  Round 238 div.1

    A:我们发现其实最后的答案只与a[i][i]有关,那么我们只记录a[i]=a[i][i],根据操作模拟就可以了。

//By BLADEVIL
#include <cstdio>
#define maxn 1010

using namespace std;

int n;
int mat[maxn][maxn],a[maxn];

int main() {
    //freopen("data.txt","r",stdin);
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++) scanf("%d",&mat[i][j]);
    for (int i=1;i<=n;i++) a[i]=mat[i][i];
    int ans(0);
    for (int i=1;i<=n;i++) ans=(ans+a[i])%2;
    int task; scanf("%d",&task);
    while (task--) {
        int x,y; scanf("%d",&x);
        if (x==3) printf("%d",ans); else {
            scanf("%d",&x); ans^=1;
        }
    }
    printf("\n");
    return 0;
}
View Code

    B:我们可以将每个给定的a[i]-1,查询s-a[i]+1位置是否被占用,被占用我们cnt++,表示和为s-1的数对加一,否则直接输出s-a[i]+1,然后再找出cnt个和为s-1的数对输出。

//By BLADEVIL
#include <cstdio>
#define maxn 1000010
#define s 1000000

using namespace std;

int n,tot;
int flag[maxn];

int main() {
    scanf("%d",&n);
    for (int i=1;i<=n;i++) {
        int x; scanf("%d",&x);
        flag[x]=1; if (flag[s+1-x]) tot++;
    }
    printf("%d\n",n);
    for (int i=1;i<=s;i++) {
        if (tot&&!flag[i]&&!flag[s-i+1]) {
            tot--;
            flag[i]=flag[s-i+1]=1;
            printf("%d %d ",i,s-i+1);
        }
        if (flag[i]&&!flag[s-i+1]) printf("%d ",s-i+1);
    }
    printf("\n");
    return 0;
}
View Code

    C:深搜,同时每个点记录这个点剩下的一条没有被选定的边,枚举这个点的所有儿子,如果儿子的没有被选定的边不为0我们就画连接儿子的边和儿子剩下的边,否则判断这个点是否有剩下的边,有的话输出连接儿子的边和剩下的边,最后返回剩下的边。

 

//By BLADEVIL
#include <cstdio>
#define maxn 100010
#define maxm 200010

using namespace std;

int n,m,l;
int pre[maxm],other[maxm],last[maxn];
int flag[maxn];

void connect(int x,int y) {
    pre[++l]=last[x];
    last[x]=l;
    other[l]=y;
}

int dfs(int x,int fa) {
    flag[x]=1; int cur(0);
    for (int p=last[x];p;p=pre[p]) {
        if (other[p]==fa) continue;
        if (flag[other[p]]==2) continue;
        int edge;
        if (!flag[other[p]]) edge=dfs(other[p],x); else edge=0;
        //printf("|%d %d\n",x,edge);
        if (edge) printf("%d %d %d\n",x,other[p],edge); else 
        if (cur) printf("%d %d %d\n",cur,x,other[p]),cur=0; else cur=other[p];
    }
    flag[x]=2;
    //printf("%d %d\n",x,cur);
    return cur;
}

int main() {
    scanf("%d%d",&n,&m);
    for (int i=1;i<=m;i++) {
        int x,y; scanf("%d%d",&x,&y);
        connect(x,y); connect(y,x);
    }
    if (m&1) printf("No solution\n\n"); else dfs(1,-1);
    return 0;
}
View Code

    D:我们将每条线段能看到的最右面的线段设为这个线段的父亲,那么一个询问的答案就是这两个点的lca,对于父亲我们可以维护一个下凸壳来计算。

//By BLADEVIL
#include <cstdio>
#include <iostream>
#include <algorithm>
#define maxn 200010
#define LL long long

using namespace std;

struct rec {
    LL x,y;
}a[maxn];

int n,t,l;
int q[maxn],jump[maxn][20];
int pre[maxn],other[maxn],last[maxn],dep[maxn];

void connect(int x,int y) {
    pre[++l]=last[x];
    last[x]=l;
    other[l]=y;
}

int lca(int x,int y) {
    if (dep[x]>dep[y]) swap(x,y);
    int det(dep[y]-dep[x]);
    for (int j=0;j<=18;j++) if (det&(1<<j)) y=jump[y][j];
    //printf("%d %d\n",x,y);
    if (x==y) return x;
    for (int j=18;j>=0;j--) 
        if (jump[x][j]!=jump[y][j]) x=jump[x][j],y=jump[y][j];
    return jump[x][0];
}

void dfs(int x) {
    for (int p=last[x];p;p=pre[p]) {
        dep[other[p]]=dep[x]+1;
        dfs(other[p]);
    }
}

int main() {
    cin>>n;
    for (int i=1;i<=n;i++) cin>>a[i].x>>a[i].y;
    for (int i=n;i;i--) {
        //for (int j=1;j<=t;j++) printf("%d ",q[j]); printf("\n");
        while (t>1&&(a[q[t]].y-a[q[t-1]].y)*(a[i].x-a[q[t-1]].x)>(a[q[t]].x-a[q[t-1]].x)*(a[i].y-a[q[t-1]].y)) t--;
        jump[i][0]=q[t];
        q[++t]=i;
    }
    //for (int i=1;i<=n;i++) printf("%d ",jump[i][0]);
    for (int i=1;i<=n;i++) connect(jump[i][0],i);
    dfs(0);
    //for (int i=1;i<=n;i++) printf("%d ",dep[i]); printf("\n");
    for (int j=1;j<=18;j++)
        for (int i=1;i<=n;i++) jump[i][j]=jump[jump[i][j-1]][j-1];
    int task; cin>>task;
    while (task--) {
        int x,y; cin>>x>>y;
        cout<<lca(x,y)<< ;
    }
    cout<<endl;
    return 0;
}
View Code

  Round 240 div.1

    A:假设n为偶数,奇数时最后一位随意,那么我们最后两个数为k-(n+2)/2和2*(k-(n+2)/2),前n-2个数为不和最后两位相同的连续的数就可以了。

//By BLADEVIL
#include <cstdio>
#define maxn 100010

using namespace std;

int n,k;
int a[maxn];

int main() {
    scanf("%d%d",&n,&k);
    if ((n==1)&&(k)) {
        printf("-1\n");
        return 0;
    }
    if (n-(n&1)>2*k) {
        printf("-1\n");
        return 0;
    }
    int cur((n-2-(n&1))/2);
    k-=cur;
    a[n]=1000001+k;
    for (int i=1,j=1;i<=n-2-(n&1);i+=2) {
        if (j==k) j+=100;
        if (j+1==k) j+=100;
        if (j==(k<<1)) j+=100;
        if (j+1==(k<<1)) j+=100;
        a[i]=j++; a[i+1]=j++;
    }
    a[n-(n&1)]=k,a[n-1-(n&1)]=k<<1;
    for (int i=1;i<=n;i++) printf(i==n?"%d\n":"%d ",a[i]);
    return 0;
}
View Code

    B:设w[i][j]为第i位为j的方案数,然后转移就可以了。

//By BLADEVIL
#include <cmath>
#include <cstdio>
#define maxn 2010
#define d39 1000000007

using namespace std;

int n,k;
int w[maxn][maxn];

int main() {
    scanf("%d%d",&n,&k);
    for (int i=1;i<=n;i++) w[1][i]=1;
    for (int i=2;i<=k;i++)
        for (int j=1;j<=n;j++)
            for (int g=1;g<=sqrt(j);g++) if (!(j%g)) {
                w[i][j]=(w[i][j]+w[i-1][g])%d39;
                if (g*g!=j) w[i][j]=(w[i][j]+w[i-1][j/g])%d39;
            }
    int ans(0);
    for (int i=1;i<=n;i++) ans=(ans+w[k][i])%d39;
    printf("%d\n",ans);
    return 0;
}
View Code

    C:我们可以将一次翻转看成将左右儿子翻转然后交换左右儿子,我们可以发现每个块内的翻转对块与块之间的答案没有影响,那么我们只需要按照归并的过程求出长度为2^j的块与块合并的时候产生的答案,和交换块产生的答案,那么每次翻转我们就可以看做将j之前的所有当前状态取反,计算代价即可。

//By BLADEVIL
#include <cstdio>
#include <cstring>
#include <iostream>
#define maxn (1<<20)+100
#define LL long long

using namespace std;

LL n,len;
LL a[maxn],b[maxn];
LL w[30][2],flag[30];

void work(LL p,LL cur) {
    if (p>n) return ; 
    for (LL i=1;i<=len;i+=cur) {
        LL k1(i),k2(i+(cur>>1)),det(0);
        while ((k1<i+(cur>>1))||(k2<i+cur)) {
            if ((k1==i+(cur>>1))||((a[k2]<a[k1])&&(k2<i+cur))) {
                w[p][0]+=i+(cur>>1)-k1;
                b[i+det]=a[k2];
                k2++;
            } else {
                b[i+det]=a[k1];
                k1++;    
            }
            det++;
        }
        k1=i+(cur>>1); k2=i;
        while ((k1<i+cur)||(k2<i+(cur>>1))) {
            if ((k1==i+cur)||((a[k2]<a[k1])&&(k2<i+(cur>>1)))) {
                w[p][1]+=i+cur-k1;
                k2++;
            } else k1++;
        }
    }
    memcpy(a,b,sizeof b);
    work(p+1,cur<<1);
}

int main() {
    cin>>n; len=1;
    for (LL i=1;i<=n;i++) len*=2;
    for (LL i=1;i<=len;i++) cin>>a[i];
    work(1,2);
    LL ans(0);
    for (LL i=1;i<=n;i++) ans+=w[i][0];
    //for (LL i=1;i<=n;i++) printf("%d ",w[i][0]); printf("\n");
    //for (LL i=1;i<=n;i++) printf("%d ",w[i][1]); printf("\n");
    for (LL i=1;i<=n;i++) flag[i]=1;
    //printf("%d\n",ans);
    LL task; cin>>task;
    while (task--) {
        LL x; cin>>x;
        for (LL i=1;i<=x;i++) ans+=w[i][flag[i]]-w[i][flag[i]^1];
        for (LL i=1;i<=x;i++) flag[i]^=1;
        cout<<ans<<endl;
    }
    return 0;
}
View Code

    Round 245 div.1

    A:tree-dp,设w[i][0..1]表示这个节点的状态和最后的相同(1),不相同(0),且每两层的儿子节点都符合这个状态的最小代价,然后转移就可以了,其实这道题相当于两棵互不影响的树,分别做tree-dp。

//By BLADEVIL
#include <cstdio>
#define maxn 100010
#define maxm 200020

using namespace std;

int n,l;
int a[maxn],b[maxn];
int other[maxm],last[maxn],pre[maxm];
int que[maxn],dep[maxn];
int w[maxn][2];

void connect(int x,int y) {
    pre[++l]=last[x];
    last[x]=l;
    other[l]=y;
}

void dfs(int x,int cur) {
    if (a[x]^b[x]^(!cur)) printf("%d\n",x),cur^=1;
    for (int p=last[x];p;p=pre[p]) if (dep[other[p]]>dep[x])
        for (int q=last[other[p]];q;q=pre[q]) if (dep[other[q]]>dep[other[p]])
            dfs(other[q],cur);
}

int main() {
    scanf("%d",&n);
    for (int i=1;i<n;i++) {
        int x,y; scanf("%d%d",&x,&y);
        connect(x,y); connect(y,x);
    }
    for (int i=1;i<=n;i++) scanf("%d",&a[i]);
    for (int i=1;i<=n;i++) scanf("%d",&b[i]);
    int h(0),t(1);
    que[1]=1; dep[1]=1;
    while (h<t) {
        int cur=que[++h];
        for (int p=last[cur];p;p=pre[p]) {
            if (dep[other[p]]) continue;
            que[++t]=other[p]; dep[other[p]]=dep[cur]+1;
        }
    }
    for (int i=n;i;i--) {
        int x=que[i];
        for (int p=last[x];p;p=pre[p]) {
            if (dep[other[p]]<dep[x]) continue;
            //w[x][0]+=w[other[p]][1]; w[x][1]+=w[other[p]][1];
            for (int q=last[other[p]];q;q=pre[q]) {
                if (dep[other[q]]<dep[other[p]]) continue;
                if (a[x]^b[x]) w[x][0]+=w[other[q]][0]; else w[x][0]+=w[other[q]][1];
                if (a[x]^b[x]) w[x][1]+=w[other[q]][0]; else w[x][1]+=w[other[q]][1];
            }
        }
        if (a[x]^b[x]) w[x][1]++; else w[x][0]++;
    }
    //for (int i=1;i<=n;i++) printf("%d %d\n",w[i][0],w[i][1]);
    int ans(w[1][1]);
    for (int p=last[1];p;p=pre[p]) ans+=w[other[p]][1];
    printf("%d\n",ans);
    dfs(1,1);
    for (int p=last[1];p;p=pre[p]) dfs(other[p],1);
    return 0;
}
View Code

    B:记录w[i][j][1..4]表示四个方向到i,j点的最大值,然后枚举相遇的节点就可以了。

//By BLADEVIL
#include <cstdio>
#include <iostream>
#include <algorithm>
#define maxn 1010
#define LL long long

using namespace std;

LL n,m;
LL a[maxn][maxn];
LL w[5][maxn][maxn];

void prepare() {
    for (LL i=1;i<=n;i++)
        for (LL j=1;j<=m;j++) w[1][i][j]=max(w[1][i][j-1],w[1][i-1][j])+a[i][j];
    for (LL i=1;i<=n;i++)
        for (LL j=m;j;j--) w[2][i][j]=max(w[2][i][j+1],w[2][i-1][j])+a[i][j];
    for (LL i=n;i;i--)
        for (LL j=1;j<=m;j++) w[3][i][j]=max(w[3][i+1][j],w[3][i][j-1])+a[i][j];
    for (LL i=n;i;i--)
        for (LL j=m;j;j--) w[4][i][j]=max(w[4][i+1][j],w[4][i][j+1])+a[i][j];
}

int main() {
    cin>>n>>m;
    for (LL i=1;i<=n;i++)
        for (LL j=1;j<=m;j++) cin>>a[i][j];
    prepare();
    LL ans(0);
    for (LL i=2;i<n;i++)
        for (LL j=2;j<m;j++) 
            ans=max(ans,w[1][i-1][j]+w[4][i+1][j]+w[2][i][j+1]+w[3][i][j-1]),
            ans=max(ans,w[1][i][j-1]+w[4][i][j+1]+w[2][i-1][j]+w[3][i+1][j]);
    cout<<ans<<endl;
    return 0;
}
View Code

    C:深搜就可以了,有一种贪心不正确但是数据弱可以过去。

#include <cstdio>
#include <iostream>
#include <algorithm>
#define maxn 30

using namespace std;

int n;
int w[maxn],p[maxn];

int dfs(int x,int used) {
    //printf("%d %d\n",x,used);
    if (x>n) return *max_element(p+1,p+1+n)==0;
    for (int i=1;i<=x;i++) 
        if ((p[i]>w[x]||p[i]==w[x]&&p[i]!=w[i]-1)&&((used&(1<<p[i])))==0) {
            p[i]-=w[x];
            if (dfs(x+1,x+1<=n&&w[x]==w[x+1]?used:0)) return 1;
            p[i]+=w[x];
            used|=1<<p[i];
        }
    return 0;
}

int main() {
    scanf("%d",&n);
    for (int i=1;i<=n;i++) scanf("%d",&w[i]);
    sort(w+1,w+1+n,greater<int>());
    for (int i=1;i<=n;i++) p[i]=w[i]-1;
    printf(dfs(2,0)?"YES\n":"NO\n");
}
View Code