HZNU 2019 Summer training 6

A - Infinite Sequence

 CodeForces - 622A 

题意：第一个数是1，接下来是1和2，接下来是1,2, 3，接下来是1,2,3, 4，问第n个数是什么

题解：找出第几轮在找出第几个

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
using namespace std;
typedef long long ll;

const int maxn = 1e2 + 10;

int main()
{
    ll n;
    while(~scanf("%lld",&n)) {
        ll tmp;
        tmp = sqrt(2 * n * 1.0);
        if(tmp * (tmp + 1) > 2 * n)
            tmp--;
        if(n - (tmp * (tmp + 1)) / 2  == 0)
            printf("%lld
",tmp);
        else
            printf("%lld
",n - (tmp * (tmp + 1)) / 2);
    }
}

 

B - Not Equal on a Segment

 CodeForces - 622C 

题意：给你一个数组，给你m个询问，每个询问一个区间， 一个数值，在区间里哪一个位置不是这个数值，输出任意一个就行

题解：找到区间里的不一样的两个数字比较一下就好了

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
using namespace std;
typedef long long ll;

const int maxn = 2e5 + 10;
int a[maxn];
struct node{
    int st,en;
    int ans;
}b[maxn];
int flag[maxn];
int main()
{
    int n, q;
    scanf("%d %d", &n, &q);
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]);
    int pos = 0;
    for (int i = 1; i <= n; i++)
    {
        if(i == 1)
        {
            b[pos].st = i;
            b[pos].ans = a[i];
        }
        else
        {
            if(a[i] != a[i-1])
            {
                b[pos].en = i - 1;
                pos++;
                b[pos].st = i;
                b[pos].ans = a[i];
            }
        }
    }
    b[pos].en = n;
    int id = 1;
    for(int i=0;i<=pos;i++)
    {
        for(int j=b[i].st;j<=b[i].en;j++)
            flag[j] = id;
        id++;
    }

//    for(int i=0;i<=pos;i++)
//        printf("%d %d %d
",b[i].st,b[i].en,b[i].ans);
//    cout<<endl;
//    for(int i=1;i<=n;i++)
//        printf("%d ",flag[i]);

    while(q--)
    {
        int l,r,x;
        scanf("%d %d %d",&l,&r,&x);
        if(flag[l] == flag[r] && a[l] == x)
        {
            puts("-1");
            continue;
        }
        else if(flag[l] == flag[r] && a[r] != x)
        {
            printf("%d
",l);
            continue;
        }
        else
        {
            int id1 = flag[l];
            id1--;
            if(b[id1].ans != x)
                printf("%d
",l);
            else
            {
                printf("%d
",b[id1 + 1].st);
            }
        }
    }

}

C - Optimal Number Permutation

 CodeForces - 622D 

题意：用1-n组长度为2n的序列，要求每个数出现2次，假设位置分别为xi、yi(xi < yi)，定义di = yi-xi。让你找到一种排列使得sigma((n-i) * |di+i-n|)最小。

题解：贪心+找规律

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
using namespace std;
typedef long long ll;

const int maxn = 2e6 + 10;
int a[maxn];
int main()
{
   int n;
   memset(a,0,sizeof a);
   scanf("%d",&n);
   int posa = 1;
   int posb = n + 1;
   for(int i=1;i<n;i++)
   {
        if(i % 2 == 1)
        {
            a[posa] = i;
            a[n + 1 - posa] = i;
            posa++;
        }
        else
        {
            a[posb] = i;
            a[3 * n - posb] = i;
            posb++;
        }
   }
   for(int i = 1;i <= 2 * n;i++)
   {
       if(a[i] == 0)
           printf("%d ",n);
       else
           printf("%d ",a[i]);

   }
}

D - Ants in Leaves

 CodeForces - 622E 

题意：给定一颗树，每个叶子节点上有一蚂蚁，除了根结点的之外的所有节点任意时刻至多只能有一个蚂蚁，每个蚂蚁每秒能移动到相邻的节点上，问所有蚂蚁移动到根结点的最短时间是多少。

题解：深度小的叶子蚂蚁先走，这样深度小的叶子蚂蚁就不用等深度大的叶子蚂蚁。然后算出每个子树走完的所需时间。d[i] = max(d[i - 1] + 1,d[i])

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
using namespace std;
typedef long long ll;

const int maxn = 5e5 + 10;

int n;
vector<int>g[maxn],d;

void dfs(int u,int fa,int deep)
{
    if(g[u].size() == 1)
    {
        d.push_back(deep);
//        printf("%d
",deep);
    }
    for(int i=0;i<g[u].size();i++)
    {
        int v = g[u][i];
        if(v == fa)
            continue;
        dfs(v,u,deep+1);
    }
}
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
            g[i].clear();
        for(int i=1;i<n;i++)
        {
            int u,v;
            scanf("%d %d",&u,&v);
            g[u].push_back(v);
            g[v].push_back(u);
        }
        int ans = 0;
        for(int k = 0; k < g[1].size(); k++)
        {
            d.clear();
            dfs(g[1][k],1,1);
            sort(d.begin(),d.end());
            for(int i = 1; i < d.size(); i++)
            {

                d[i] = max(d[i - 1] + 1,d[i]);
                //cout<<d[i]<<endl;
            }
            ans = max(ans,d.back());
        }
        printf("%d
",ans);
    }
    return 0;
}

E - The Sum of the k-th Powers

 CodeForces - 622F 

题意：求   mod (1e9 + 7)  

 题解：朗格朗日差值板子 

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#include<string.h>
#include<cstring>
using namespace std;
#define LL long long
const int MAXN = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod =  1e9 + 7;

LL n,k,ans,fac[MAXN],y[MAXN];

LL qpow(LL a, LL b)
{
    LL s = 1;
    while(b)
    {
        if(b & 1)
            s = s * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return s;
}
void init()
{
    fac[0] = fac[1] = y[1] = 1;
    for(int i = 2; i < MAXN; i++)
        fac[i] = fac[i - 1] * i % mod;
    for(int i = 2; i < MAXN; i++)
        y[i] = (y[i - 1] + qpow(i,k)) % mod;
}
int main()
{
    cin >> n >> k;
    init();
    if(k == 0)
        cout << n << endl;
    else if(n <= k + 2)
    {
        cout << y[n] << endl;
    }
    else
    {
        LL sum = 1,sig;
        for(LL i = n - k - 2; i <= n - 1; i++)
            sum = sum * i % mod;
        for(LL i = 1; i <= k + 2; i++)
        {
            LL fz = sum * qpow(n - i,mod - 2) % mod;
            LL fm = qpow(fac[i - 1] * fac[k + 2 - i] % mod,mod - 2);
            if((k + 2 - i) % 2 == 0)
                sig = 1;
            else
                sig = -1;
            ans += sig * y[i] * fz % mod * fm % mod;
            ans %= mod;
        }
        cout << (ans % mod + mod) % mod << endl;
    }
}