2018多校联合训练1

http://acm.hdu.edu.cn/contests/contest_show.php?cid=802

三人三台电脑比赛，然而有个队友鸽了,只有两人打，一共过了5题

1001：水题，只需判断3和4的倍数即可，solved by lyy

#include 
using namespace std;
#define ll long long
int t,n;

int main()
{
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d",&n);
        if (n%3==0) printf("%lld\n",(ll)(n/3)*(n/3)*(n/3));
        else if (n%4==0) printf("%lld\n",2LL*(n/4)*(n/4)*(n/4));
        else printf("-1\n");
    }
    return 0;
}

 

1003：水题，排序之后顺序取点即可, solved by lyy

#include 
using namespace std;
#define ll long long
int t,n;
struct point
{
    int id;
    int x,y;
}p[3005];

bool cmp(point a,point b)
{
    return a.x

 

1011：字符串处理，solved by sdn

/*
  ID: oodt
  PROG:
  LANG:C++
*/
#include
#include
#include
#include
#include
#include
#include

 

1004：排序之后，直接暴力往数组里填数即可， solved by lyy

#include 
using namespace std;
#define ll long long
int t,n,m;
int a[100005];
int sum[100005];
struct st
{
    int l,r;
}p[100005];

bool cmp(st a,st b)
{
    if (a.l!=b.l) return a.lr)
            {
                for (int j=l;j<=r;j++)
                {
                    sum[a[j]]=0;
                    l++;
                }
                for (int j=l;jr)
                {
                    for (int j=r;j0) mi++;
                    }
                }
            }
        }
        int ma=p[1].r;
        for (int i=1;i<=m;i++)
        {
            ma=max(ma,p[i].r);
        }
        for (int i=ma+1;i<=n;i++)
        {
            a[i]=1;
        }
        for (int i=1;i<=n;i++)
        {
            printf("%d",a[i]);
            if (i!=n) printf(" ");
        }
        printf("\n");
    }
    return 0;
}

 

1007：打表，然后扔到OEIS里，找到一段python代码，研究一下发现可以二分写, solved by lyy

#include 
using namespace std;
#define ll long long
const ll mod=1000000007;
int t;
ll n;

long long inv(long long a,long long m)
{
    if(a == 1)return 1;
    return inv(m%a,m)*(m-m/a)%m;
}

ll inv2=inv(2,mod);

ll f(ll x)
{
    if (x==1) return 1;
    else return x+f(x/2);
}

ll find(ll n)
{
    ll l=1,r=n;
    ll m;
    while (l<=r)
    {
        m=(l+r)/2;
        if (f(m)