2017.10.28 QB模拟赛 —— 上午

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);
        ,r=;
        for(long long mid;l<=r;)
        {
            mid=(l+r)>>;
            ;
            ;
            else {puts("YES");goto flag;}
        }
        puts("NO");
        flag:;
    }
    fclose(stdin); fclose(stdout);
    ;
}

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);
        ,r=;
        for(long long mid;l<=r;)
        {
            mid=(l+r)>>;
            )+(mid+)*(mid+)>p) r=mid-;
            )+(mid+)*(mid+)<p) l=mid+;
            else {puts("YES");goto flag;}
        }
        puts("NO");
        flag:;
    }
    fclose(stdin); fclose(stdout);
    ;
}

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();
    ;!isdigit(ch);ch=getchar());
    +ch-',ch=getchar());
}
|],val[N<<|];
void build(int k,int l,int r)
{
    flag[k]=;
    ;return;}
    ;
    build(k<<,l,mid);
    build(k<<|,mid+,r);
    val[k]=min(val[k<<],val[k<<|]);
}
void down(int k)
{
    val[k<<]=flag[k];
    val[k<<|]=flag[k];
    flag[k<<]=flag[k];
    flag[k<<|]=flag[k];
    flag[k]=;
}
int query(int k,int l,int r,int x,int y)
{
    if(l>=x&&r<=y) return val[k];
    ,ret=0x7fffffff;
    if(flag[k]) down(k);
    ,l,mid,x,y));
    |,mid+,r,x,y));
    val[k]=min(val[k<<],val[k<<|]);
    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;}
    ;
    if(flag[k]) down(k);
    ,l,mid,x,y,v);
    |,mid+,r,x,y,v);
    val[k]=min(val[k<<],val[k<<|]);
}
bool check(int k)
{
    build(,,n);
    ;i<=k;++i) b[i]=a[i];
    sort(b+,b++k);
    int lmax,lmin,rmax,rmin;
    ;i<=k;)
    {
        int j,x=b[i].x;
        lmax=b[i].l,rmax=b[i].r,lmin=b[i].l,rmin=b[i].r;
        ;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;
        }
        ,,n,lmin,rmax);
        if(re) return true;
        modify(,,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);
    ;i<=t;++i)
    {
        read(a[i].l); read(a[i].r); read(a[i].x);
        a[i].t=i;
    }
    ,l=,r=t;
    for(int mid;l<=r;)
    {
        mid=(l+r)>>;
        ;
        ;
    }
    printf("%d",ans);
    fclose(stdin); fclose(stdout);
    ;
}

#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()
{
    , f = ;
    char ch = getchar();
    ;
    ) + (x << ) + ch - ';
    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;
    ; i <= n + ; i++) f[i] = i;
    ; i <= k; i++) t[i] = p[i];
    std::sort(t + , t + k + , cmp);
    lmin = lmax = t[].x;
    rmin = rmax = t[].y;
    ; i <= k; i++)
    {
        ].z)
        {
            ;
            for(j = find(lmin); j <= rmax; j++)
                f[find(j)] = find(rmax + );
            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);
            ;
        }
    }
//    cout<<find(1)<<endl;
    ;
    ;
}
 
int main()
{
    freopen("number.in","r",stdin);
    freopen("number.out","w",stdout);
    int i, x, y, mid;
    n = read();
    q = read();
    ; i <= q; i++)
        p[i].x = read(), p[i].y = read(), p[i].z = read();
    x = , y = q;
    //cout<<check(2)<<endl;
    //return 0;
    ans = q + ;
    while(x <= y)
    {
        mid = (x + y) >> ;
        ;
        ;
    }
    printf("%d\n", ans);
    ;
}

std并查集维护