2017.10.28 QB模拟赛 ——上午

Posted 2020-10-13

题目链接

T1

1e18 内的立方数有 1e6个 

直接枚举可过 

二分最优 

考场用set  死慢。。

#include <cstdio>
int t;
long long p;
int main(int argc,char *argv[])
{
    freopen("cubic.in","r",stdin);
    freopen("cubic.out","w",stdout);
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d",&p);
        long long l=1,r=1000000;
        for(long long mid;l<=r;)
        {
            mid=(l+r)>>1;
            if(mid*mid*mid>p) r=mid-1;
            else if(mid*mid*mid<p) l=mid+1;
            else {puts("YES");goto flag;} 
        }
        puts("NO");
        flag:;
    }
    fclose(stdin); fclose(stdout);
    return 0;
}

View Code

 

T2

p=a^3-b^3

由立方差公式 a^3-b^3=(a-b)*(a^2+ab+b^2)

因为p是素数 所以 a-b=1;

二分即可 

#include <cstdio>
int t;
long long p;
int main(int argc,char *argv[])
{
    freopen("cubicp.in","r",stdin);
    freopen("cubicp.out","w",stdout);
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d",&p);
        long long l=1,r=1000000;
        for(long long mid;l<=r;)
        {
            mid=(l+r)>>1;
            if(mid*mid+mid*(mid+1)+(mid+1)*(mid+1)>p) r=mid-1;
            else if(mid*mid+mid*(mid+1)+(mid+1)*(mid+1)<p) l=mid+1;
            else {puts("YES");goto flag;}
        }
        puts("NO");
        flag:;
    }
    fclose(stdin); fclose(stdout);
    return 0;
}

View Code

 

T3

二分答案 

线段树维护区间

#include <algorithm>
#include <cctype>
#include <cstdio>
#define N 1000005
using namespace std;

struct node
{
    int l,r,x,t;
    bool operator<(node a)const
    {
        return x>a.x;
    }
}a[N],b[N];
inline void read(int &x)
{
    register char ch=getchar();
    for(x=0;!isdigit(ch);ch=getchar());
    for(;isdigit(ch);x=x*10+ch-‘0‘,ch=getchar());
}
int n,t,flag[N<<2|1],val[N<<2|1];
void build(int k,int l,int r)
{
    flag[k]=0;
    if(l==r) {val[k]=0;return;}
    int mid=(l+r)>>1;
    build(k<<1,l,mid);
    build(k<<1|1,mid+1,r);
    val[k]=min(val[k<<1],val[k<<1|1]);
}
void down(int k)
{
    val[k<<1]=flag[k];
    val[k<<1|1]=flag[k];
    flag[k<<1]=flag[k];
    flag[k<<1|1]=flag[k];
    flag[k]=0;
}
int query(int k,int l,int r,int x,int y)
{
    if(l>=x&&r<=y) return val[k];
    int mid=(l+r)>>1,ret=0x7fffffff;
    if(flag[k]) down(k);
    if(x<=mid) ret=min(ret,query(k<<1,l,mid,x,y));
    if(y>mid) ret=min(ret,query(k<<1|1,mid+1,r,x,y));
    val[k]=min(val[k<<1],val[k<<1|1]);
    return ret;
}
void modify(int k,int l,int r,int x,int y,int v)
{
    if(l>=x&&r<=y) {flag[k]=val[k]=v;return;}
    int mid=(l+r)>>1;
    if(flag[k]) down(k);
    if(x<=mid) modify(k<<1,l,mid,x,y,v);
    if(y>mid) modify(k<<1|1,mid+1,r,x,y,v);
    val[k]=min(val[k<<1],val[k<<1|1]);
}
bool check(int k)
{
    build(1,1,n);
    for(int i=1;i<=k;++i) b[i]=a[i];
    sort(b+1,b+1+k);
    int lmax,lmin,rmax,rmin;
    for(int i=1;i<=k;)
    {
        int j,x=b[i].x;
        lmax=b[i].l,rmax=b[i].r,lmin=b[i].l,rmin=b[i].r;
        for(j=i+1;b[j].x==x&&j<=k;++j)
        {
            lmax=max(lmax,b[j].l);
            lmin=min(lmin,b[j].l);
            rmin=min(rmin,b[j].r);
            rmax=max(rmax,b[j].r);
            if(lmax>rmin) return true;
        }
        int re=query(1,1,n,lmin,rmax);
        if(re) return true;
        modify(1,1,n,lmin,rmax,b[i].x); i=j;
    }
    return false;
}
int main(int argc,char *argv[])
{
    freopen("number.in","r",stdin);
    freopen("number.out","w",stdout);
    read(n); read(t);
    for(int i=1;i<=t;++i)
    {
        read(a[i].l); read(a[i].r); read(a[i].x);
        a[i].t=i;
    }
    int ans=t+1,l=1,r=t;
    for(int mid;l<=r;)
    {
        mid=(l+r)>>1;
        if(check(mid)) ans=mid,r=mid-1;
        else l=mid+1;
    }
    printf("%d",ans);
    fclose(stdin); fclose(stdout);
    return 0;
}

View Code

#include <cstdio>
#include <iostream>
#include <algorithm>
#define N 1000011
#define min(x, y) ((x) < (y) ? (x) : (y))
#define max(x, y) ((x) > (y) ? (x) : (y))
using namespace std;
int n, q, ans;
int f[N];

struct node
{
    int x, y, z;
}p[N], t[N];

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

inline bool cmp(node x, node y)
{
    return x.z > y.z;
}

inline int find(int x)
{
    return x == f[x] ? x : f[x] = find(f[x]);
}

inline bool check(int k)
{
    int i, j, x, y, lmin, lmax, rmin, rmax;
    for(i = 1; i <= n + 1; i++) f[i] = i;
    for(i = 1; i <= k; i++) t[i] = p[i];
    std::sort(t + 1, t + k + 1, cmp);
    lmin = lmax = t[1].x;
    rmin = rmax = t[1].y;
    for(i = 2; i <= k; i++)
    {
        if(t[i].z < t[i - 1].z)
        {
            if(find(lmax) > rmin) return 1;
            for(j = find(lmin); j <= rmax; j++)
                f[find(j)] = find(rmax + 1);
            lmin = lmax = t[i].x;
            rmin = rmax = t[i].y;
        }
        else
        {
            lmin = min(lmin, t[i].x);
            lmax = max(lmax, t[i].x);
            rmin = min(rmin, t[i].y);
            rmax = max(rmax, t[i].y);
            if(lmax > rmin) return 1;
        }
    }
//    cout<<find(1)<<endl;
    if(find(lmax) > rmin) return 1;
    return 0;
}

int main()
{
    freopen("number.in","r",stdin);
    freopen("number.out","w",stdout);
    int i, x, y, mid;
    n = read();
    q = read();
    for(i = 1; i <= q; i++)
        p[i].x = read(), p[i].y = read(), p[i].z = read();
    x = 1, y = q;
    //cout<<check(2)<<endl;
    //return 0;
    ans = q + 1;
    while(x <= y)
    {
        mid = (x + y) >> 1;
        if(check(mid)) ans = mid, y = mid - 1;
        else x = mid + 1;
    }
    printf("%d\n", ans);
    return 0;
}

std并查集维护