2017-10-湖南套题1

复杂度m*logn是可以通过的，可以考虑每次询问时二分查找

题目要求所有的线段没有交点，那么肯定是最靠近原点的两点连线，次靠近的。。。像这样：最下面是①号线，向上递增

对于每次询问的(px,py),可以发现，在x=px这条直线上，对于每条线段的y是单调递增的，就可以二分是第几条线段的y>=px

第i条可以，那么[1,i]条均可以，即可以有i个交点

#include <algorithm>
#include <cstdio>

#define dou double

const int N(1e5+5);
struct ZhiXian {
    double k,b;
}y[N];

dou a[N],b[N],px,py;
int n,m,ans[N];

int L,R,Mid;
inline bool check(int x)
{
    return (px*y[x].k+y[x].b)<=py;
}
inline int erfen()
{
    int ret=0;
    for(L=0,R=n; L<=R; )
    {
        Mid=L+R>>1;
        if(check(Mid))
        {
            ret=Mid;
            L=Mid+1;
        }
        else R=Mid-1;
    }
    return ret;
}

int Presist()
{
    freopen("geometry.in","r",stdin);
    freopen("geometry.out","w",stdout);
    scanf("%d",&n);
    for(int i=1; i<=n; ++i) scanf("%lf",a+i);
    for(int i=1; i<=n; ++i) scanf("%lf",b+i);
    std::sort(a+1,a+n+1);std::sort(b+1,b+n+1);
    for(int i=1; i<=n; ++i)
        y[i].b=b[i],y[i].k=-b[i]/a[i];
    scanf("%d",&m);
    for(; m--; )
    {
        scanf("%lf%lf",&px,&py);
        printf("%d
",erfen());
    }
    return 0;
}

int Aptal=Presist();
int main(int argc,char**argv){;}

#include <cstdio>

inline void read(int &x)
{
    x=0; register char ch=getchar();
    for(; ch>'9'||ch<'0'; ) ch=getchar();
    for(; ch>='0'&&ch<='9'; ch=getchar()) x=x*10+ch-'0';
}
const int N(24);
int n,a[N],b[N];

int Presist()
{
//    freopen("cashier.in","r",stdin);
//    freopen("cashier.out","w",stdout);
    int t; read(t);
    for(int i,j,k,x,y; t--; )
    {
        int ans=0;
        for(i=0; i<24; ++i) read(a[i]);
        for(i=0; i<24; ++i) read(b[i]);
        for(x,y,j,k,i=0; i<24; ++i)
        {
            for(j=i,x=0; a[i]>0&&++x<8; --j)
            {
                if(j<0) j+=N;
                if(!b[j]) continue;
                if(b[j]>=a[i])
                {
                    ans+=a[i];b[j]-=a[i];
                    for(k=(j+1)%N,y=0; ++y<8; ++k,k%=N)
                        if(k!=i) a[k]-=a[i]; a[i]=0;
                }
                else
                {
                    ans+=b[j]; a[i]-=b[j];
                    for(k=(j+1)%N,y=0; ++y<8; ++k,k%=N)
                        if(k!=i) a[k]-=b[j];
                }
            }
            if(a[i]>0) { ans=-1; goto print; }
        }
        /*for(x,y,j,k,i=0; i<8; ++i)
        {
            for(j=i,x=0; a[i]>0&&++x<8; --j,j+=N,j%=N)
            {
                if(!b[j]) continue;
                if(b[j]>=a[i])
                {
                    ans+=a[i];b[j]-=a[i];
                    for(k=j+1,y=0; ++y<8; ++k,k+=N,k%=N)
                        if(k!=i) a[k]-=a[i]; a[i]=0;
                }
                else
                {
                    ans+=b[j]; a[i]-=b[j];
                    for(k=j+1,y=0; ++y<8; ++k,k+=N,k%=N)
                        if(k!=i) a[k]-=b[j];
                }
            }
            if(a[i]>0) { ans=-1; goto print; }
        }*/
        print:printf("%d
",ans);
    }
    return 0;
}

int Aptal=Presist();
int main(int argc,char**argv){;}

 差分约束，

设sum[i]表示使用的最小的b的前缀和，则ans=sum[23]

sum[i]需要满足 (i>7) sum[i]-sum[i-8]>=a[i] ,0<=sum[i]-sum[i-1]<=b[i]（i==0是特判向源点连）

　　　　　　　　(i<8) sum[23]-sum[i+16]+sum[i]>=a[i]

求最小值出现负环表示无解

#include <cstring>
#include <cstdio>
#include <queue>

#define max(a,b) (a>b?a:b)

inline void read(int &x)
{
    x=0; register char ch=getchar();
    for(; ch>'9'||ch<'0'; ) ch=getchar();
    for(; ch>='0'&&ch<='9'; ch=getchar()) x=x*10+ch-'0';
}
const int N(24);
int n,a[N],b[N];

int head[N+5],sumedge;
struct Edge {
    int v,next,w;
    Edge(int v=0,int next=0,int w=0):v(v),next(next),w(w){;}
}edge[N*N];

inline void ins(int u,int v,int w)
{
    edge[++sumedge]=Edge(v,head[u],w); head[u]=sumedge;
}

bool inq[N+5];
int sum[N+5],cnt[N+5];
bool SPFA()
{
    std::queue<int>que;
    for(int i=0; i<N; ++i)
        inq[i]=cnt[i]=0,sum[i]=- 0x7fffffff;
    sum[N]=0;     que.push(N);
    for(int u,v; !que.empty(); )
    {
        u=que.front(); que.pop(); inq[u]=0;
        for(int i=head[u]; i; i=edge[i].next)
        {
            v=edge[i].v;
            if(sum[v]<sum[u]+edge[i].w)
            {
                sum[v]=sum[u]+edge[i].w;
                if(++cnt[v]>N) return 0;
                if(!inq[v]) inq[v]=1,que.push(v);
            }
        }
    }
    return 1;
}

int L,R,Mid;
inline bool check(int x)
{
    memset(edge,0,sizeof(edge));
    memset(head,0,sizeof(head));
    sumedge=0;
    for(int i=0; i<24; ++i)
    {
        if(i>=8) ins(i-8,i,a[i]);
        else ins(i+16,i,a[i]-x);
        if(i) ins(i-1,i,0),ins(i,i-1,-b[i]);
        else ins(N,i,0),ins(i,N,-b[i]);
    }
    ins(N,23,x),ins(23,N,-x);
    return SPFA();
}

int Presist()
{
    freopen("cashier.in","r",stdin);
    freopen("cashier.out","w",stdout);
    int t; read(t);
    for(int i,j,k,x,y; t--; )
    {
        int ans=-1;L=0,R=0;
        for(i=0; i<24; ++i) read(a[i]),L=max(L,a[i]);
        for(i=0; i<24; ++i) read(b[i]),R+=b[i];
        for(; L<=R; )
        {
            Mid=L+R>>1;
            if(check(Mid))
            {
                R=Mid-1;
                ans=Mid;
            }
            else L=Mid+1;
        }
        printf("%d
",ans);
    }
    return 0;
}

int Aptal=Presist();
int main(int argc,char**argv){;}

 

 

 

 

#include <cstdio>

#define LL long long
const int mod(1e9+7);
const int N(50005);
long long val[N];

int Presist()
{
    freopen("game.in","r",stdin);
    freopen("game.out","w",stdout);
    int n,m;    scanf("%d%d",&n,&m);
    for(int i=1; i<=n; ++i) scanf("%I64d",val+i);
    for(int oo,a,b,c; m--; )
    {
        scanf("%d%d%d",&oo,&a,&b);
        if(oo==1)
        {
            scanf("%d",&c);
            for(int i=a; i<=b; ++i)
            {
                val[i]+=c;
                for(; val[i]<0; ) val[i]+=mod;
                for(; val[i]>=mod; ) val[i]-=mod;
            }
        }
        else if(oo==2)
            for(int i=a; i<=b; ++i)
            {
                val[i]*=-1;
                for(; val[i]<0; ) val[i]+=mod;
                for(; val[i]>=mod; ) val[i]-=mod;
            }
        else
        {
            scanf("%d",&c);LL ans=0;
            for(int i=a; i<=b; ++i)
            {
                ans+=val[i];
                for(; ans<0; ) ans+=mod;
                for(; ans>=mod; ) ans-=mod;
            }
            printf("%I64d
",ans);
        }
    }
    return 0;
}

int Aptal=Presist();
int main(int argc,char**argv){;}

k比较小，所以线段树每个节点维护一个区间答案记为f[i]

考虑一段区间i,左边取j个右边就取i-j个 答案是每个方案的左边乘右边的和。

就是i左儿子f[j]和右边的f[i-j] 所以f[i]=Σ(j=0~i) lc f[j]*rc f[i-j]

考虑取反操作，i是奇数就取反，偶数无影响(因为是相乘)

考虑区间加， 开始f[i] 是 a1*a2……an  后来是(a1+c)*(a2+c)……(an+c)

考虑类似二项式定理,当上述a1~an  n个方案如果取了j个了，分别为al1,al2……alj

那考虑最后答案，如果已经选了j个方案是(al1+c)(al2+c)……(alj+c)再还能选i-j个 最后答案是C(len-i,i-j)*f[j]*c^(i-j)

复杂度 O(k^2*nlogn)

#include <cstdio>
#include <cstdlib>
#define MOD 1000000007
#define N 100005
typedef long long LL;
using namespace std;
struct Node
{
    LL f[11];
} node[N * 4];
LL a[N], lazy1[N * 4];
bool lazy2[N * 4];
LL C[N][11];

Node merge(Node lc, Node rc)
{
    Node o;
    o.f[0] = 1;
    for (int i = 1; i <= 10; i++)
    {
        o.f[i] = 0;
        for (int j = 0; j <= i; j++)
            o.f[i] = (o.f[i] + lc.f[j] * rc.f[i - j] % MOD) % MOD;
    }
    return o;
}

void build(int o, int l, int r)
{
    if (l == r)
    {
        for (int i = 0; i <= 10; i++) node[o].f[i] = 0;
        node[o].f[0] = 1;
        node[o].f[1] = (a[l] % MOD + MOD) % MOD;
        return ;
    }
    int mid = (l + r) >> 1;
    build(o * 2, l, mid);
    build(o * 2 + 1, mid + 1, r);
    node[o] = merge(node[o * 2], node[o * 2 + 1]);
    return ;
}

void update1(int o, int l, int r, int c)
{
    int len = r - l + 1;
    LL ff[11];
    for (int i = 0; i <= 10; i++) ff[i] = node[o].f[i];
    for (int i = 1; i <= 10; i++)
    {
        node[o].f[i] = 0;
        LL t = 1;
        for (int j = 0; j <= i; j++)
        {
            LL tmp = ff[i - j] * C[len - (i - j)][j] % MOD * t % MOD;
            node[o].f[i] = (node[o].f[i] + tmp) % MOD;
            t = t * c % MOD;
        }
    }
    return ;
}

void push_down(int o, int l, int r)
{
    int mid = (l + r) >> 1;
    if (lazy1[o])
    {
        if (lazy2[o * 2])
            lazy1[o * 2] = (lazy1[o * 2] + MOD - lazy1[o]) % MOD;
        else
            lazy1[o * 2] = (lazy1[o * 2] + lazy1[o]) % MOD;
        if (lazy2[o * 2 + 1])
            lazy1[o * 2 + 1] = (lazy1[o * 2 + 1] + MOD - lazy1[o]) % MOD;
        else
            lazy1[o * 2 + 1] = (lazy1[o * 2 + 1] + lazy1[o]) % MOD;
        update1(o * 2, l, mid, lazy1[o]);
        update1(o * 2 + 1, mid + 1, r, lazy1[o]);
        lazy1[o] = 0;
    }
    if (lazy2[o])
    {
        lazy2[o * 2] ^= 1;
        lazy2[o * 2 + 1] ^= 1;
        for (int j = 1; j <= 10; j += 2)
        {
            node[o * 2].f[j] = MOD - node[o * 2].f[j];
            node[o * 2 + 1].f[j] = MOD - node[o * 2 + 1].f[j];
        }
        lazy2[o] = 0;
    }
}

void modify1(int o, int l, int r, int ll, int rr, int c)
{
    if (ll <= l && rr >= r)
    {
        if (lazy2[o]) lazy1[o] = (lazy1[o] + MOD - c) % MOD;
        else lazy1[o] = (lazy1[o] + c) % MOD;
        update1(o, l, r, c);
        return ;
    }
    int mid = (l + r) >> 1;
    push_down(o, l, r);
    if (ll <= mid) modify1(o * 2, l, mid, ll, rr, c);
    if (rr > mid) modify1(o * 2 + 1, mid + 1, r, ll, rr, c);
    node[o] = merge(node[o * 2], node[o * 2 + 1]);
    return ;
}

void modify2(int o, int l, int r, int ll, int rr)
{
    if (ll <= l && rr >= r)
    {
        for (int i = 1; i <= 10; i += 2) node[o].f[i] = MOD - node[o].f[i];
        lazy2[o] ^= 1;
        return ;
    }
    int mid = (l + r) >> 1;
    push_down(o, l, r);
    if (ll <= mid) modify2(o * 2, l, mid, ll, rr);
    if (rr > mid) modify2(o * 2 + 1, mid + 1, r, ll, rr);
    node[o] = merge(node[o * 2], node[o * 2 + 1]);
    return ;
}

Node query(int o, int l, int r, int ll, int rr)
{
    if (ll <= l && rr >= r)
        return node[o];
    int mid = (l + r) >> 1;
    push_down(o, l, r);
    if (rr <= mid) return query(o * 2, l, mid, ll, rr);
    if (ll > mid) return query(o * 2 + 1, mid + 1, r, ll, rr);
    Node lc = query(o * 2, l, mid, ll, rr);
    Node rc = query(o * 2 + 1, mid + 1, r, ll, rr);
    return merge(lc, rc);
}

int main(int argc, char ** argv)
{
    freopen("game.in", "r", stdin);
    freopen("game.out", "w", stdout);
    int n, m;
    scanf("%d %d", &n, &m);
    C[0][0] = 1;
    for (int i = 1; i <= n; i++)
    {
        C[i][0] = 1;
        for (int j = 1; j <= 10; j++)
            C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD;
    }
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]);
    build(1, 1, n);
    for (int i = 1; i <= m; i++)
    {

        int l, r, opt;
        scanf("%d%d%d",&opt, &l, &r);
        if (opt == 1)
        {
            int c;
            scanf("%d", &c);
            c = (c % MOD + MOD) % MOD;
            modify1(1, 1, n, l, r, c);
        }
        else if (opt == 2)
        {
            modify2(1, 1, n, l, r);
        }
        else
        {
            int k;
            scanf("%d", &k);
            Node o = query(1, 1, n, l, r);
            printf("%d
", o.f[k] % MOD);
        }
    }
    return 0;
}