2018/7/20 模拟赛

T1 以前做过的原题 GSS3。线段树。

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>

using namespace std;
const int MAXN = 500005;

inline int rd(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch))  {if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))  {x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
    return x*f;
}

int n,m,a[MAXN];
struct Node{
    int mx,lx,rx,sum;
}node[MAXN<<2];

inline void pushup(int x){
    node[x].sum=node[x<<1].sum+node[x<<1|1].sum;
    node[x].lx=max(node[x<<1].lx,node[x<<1].sum+node[x<<1|1].lx);
    node[x].rx=max(node[x<<1|1].rx,node[x<<1|1].sum+node[x<<1].rx);
    node[x].mx=max(node[x<<1|1].mx,max(node[x<<1].mx,node[x<<1].rx+node[x<<1|1].lx));
}

inline void build(int x,int l,int r){
    if(l==r){
        node[x].sum=node[x].lx=node[x].rx=node[x].mx=a[l];
        return ;
    }
    int mid=l+r>>1;
    build(x<<1,l,mid);build(x<<1|1,mid+1,r);
    pushup(x);
}

inline Node query(int x,int l,int r,int L,int R){
    if(L<=l && r<=R) return node[x];
    int mid=l+r>>1;
    if(mid<L) return query(x<<1|1,mid+1,r,L,R);
    if(mid>=R) return query(x<<1,l,mid,L,R);
    else{
        Node A=query(x<<1,l,mid,L,R);
        Node B=query(x<<1|1,mid+1,r,L,R);
        Node ans;
        ans.sum=A.sum+B.sum;
        ans.lx=max(A.lx,A.sum+B.lx);
        ans.rx=max(B.rx,B.sum+A.rx);
        ans.mx=max(max(A.mx,B.mx),A.rx+B.lx);
        return ans;
    }
}

inline void update(int x,int l,int r,int L,int R,int k){
    if(l==r && l==L){
        node[x].sum=node[x].mx=node[x].lx=node[x].rx=k;
        return;
    }
    int mid=l+r>>1;
    if(mid>=L) update(x<<1,l,mid,L,R,k);
    if(mid<R)  update(x<<1|1,mid+1,r,L,R,k);
    pushup(x);
}

int main(){
    freopen("BRS.in","r",stdin);
    freopen("BRS.out","w",stdout);
    n=rd();m=rd();
    for(register int i=1;i<=n;i++) a[i]=rd();
    build(1,1,n);
    for(register int i=1;i<=m;i++){
        int op=rd(),x=rd(),y=rd();
        if(op==1)   printf("%d
",query(1,1,n,x,y).mx);
        else update(1,1,n,x,x,y);
    }
    return 0;
}

T2打表找规律，最后发现是一个类似斐波那契的，矩阵乘法直接搞。结果考场上最后一步没有取模爆零？？？？？233333.。。

代码

#include<iostream> 
#include<cstdio>
#include<cstring>
#include<cmath>
#define LL long long

using namespace std;
const int mod = 7777777;

int n,k;

struct Mat{
    LL a[15][15];
    Mat(){
        memset(a,0,sizeof(a));
    }
    Mat operator*(const Mat &h){
        Mat c;
        for(register int i=1;i<=k;i++)
            for(register int j=1;j<=k;j++)
                for(register int o=1;o<=k;o++)
                    (c.a[i][j]+=(a[i][o]%mod*h.a[o][j]%mod))%=mod;
        return c;
    }
}f,ans,st;

inline void fast_pow(Mat A,int b){
    for(;b;b>>=1){
        if(b&1) st=st*A;
        A=A*A;
    }
}

int main(){
    freopen("fyfy.in","r",stdin);
    freopen("fyfy.out","w",stdout);
    scanf("%d%d",&k,&n);
    if(k==1){
        puts("1");
        return 0;
    }
    for(register int i=2;i<=k;i++)  f.a[i][i-1]=1;
    for(register int i=1;i<=k;i++)  f.a[i][k]=1;
    for(register int i=1;i<=k;i++)  st.a[i][i]=1;
    fast_pow(f,n);
    printf("%lld
",st.a[k][k]);
    return 0;
}

T3，扫描线。。当时的坑没填，暴力30分走人。先把纵坐标离散化，之后将横坐标排序，然后一个个的跳。因为数据范围小，直接O（n^2） 能过，就没线段树优化。

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
#define int long long 

using namespace std;
const int MAXN = 105;

inline int rd(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
    while(isdigit(ch))  {x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
    return x*f;
}

int n;

struct Node{
    int x1,x2,y1,y2;
}node[MAXN];

struct A{
    int y1,y2,x,k;
}a[MAXN<<1];

int y[MAXN<<1],cnt;
int c[MAXN<<1];
LL ans;

inline bool cmp(A a,A b){
    return a.x<b.x;
}

signed main(){
    freopen("olddriver.in","r",stdin);
    freopen("olddriver.out","w",stdout);
    n=rd();
    for(register int i=1;i<=n;i++){
        node[i].x1=rd(),node[i].y1=rd();
        node[i].x2=rd(),node[i].y2=rd();
        y[++cnt]=node[i].y1;
        y[++cnt]=node[i].y2;
    }
    sort(y+1,y+1+cnt);
    int u=unique(y+1,y+1+cnt)-y-1;
    for(register int i=1;i<=n;i++){
        node[i].y1=lower_bound(y+1,y+1+u,node[i].y1)-y;
        node[i].y2=lower_bound(y+1,y+1+u,node[i].y2)-y;
    }
//  for(register int i=1;i<=n;i++)
//      cout<<node[i].y1<<" "<<node[i].y2<<endl;
    cnt=0;
    for(register int i=1;i<=n;i++){
        a[++cnt].y1=node[i].y1;
        a[cnt].y2=node[i].y2;
        a[cnt].x=node[i].x1;
        a[cnt].k=1;
        a[++cnt].y1=node[i].y1;
        a[cnt].y2=node[i].y2;
        a[cnt].x=node[i].x2;
        a[cnt].k=-1;
    }
    sort(a+1,a+1+cnt,cmp);
    for(register int i=1;i<=cnt;i++){
        for(register int j=1;j<=u;j++)
            if(c[j]) ans+=(a[i].x-a[i-1].x)*(y[j+1]-y[j]);
        for(register int j=a[i].y1;j<a[i].y2;j++)
             c[j]+=a[i].k;
    }
    printf("%lld",ans);
    return 0;
}