HZNU 2019 Summer training 4

A - Little C Loves 3 I

 CodeForces - 1047A 

题意：一个数分成三份，每一个都不是3的倍数

题解：分成 (1, 1, n - 2) 或者分成(1, 2, n - 3 )

#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 = 1e3+7;

int main()
{
    ll n;
    scanf("%lld",&n);
    if((n - 2) % 3 == 0)
        printf("1 2 %lld
",n - 3);
    else
        printf("1 1 %lld
",n - 2);
}

 

B - Cover Points

 CodeForces - 1047B 

题意：平面上有n个点，用一个顶点在原点，两直角边分别在x轴和y轴的等腰直角三角形覆盖这些点，问能将这些点全部覆盖的三角形的直角边最短是多长

题解：求解max(x + y)

#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 = 1e3+7;

int main()
{
    int n;
    scanf("%d",&n);
    int maxx = 0;
    while(n--)
    {
        int a,b;
        scanf("%d %d",&a,&b);
        maxx = max(maxx,a + b);
    }
    printf("%d
",maxx);
}

 

C - Enlarge GCD

 CodeForces - 1047C 

题意：n个数的gcd是k，要你删掉最少的数使得删完后的数组的gcd > k

题解：先求出k，然后每个数除以k。然后找出出现次数最多的质因数即可。

#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 = 3e5 + 10;
const int M = 1.5e7 + 10;
int pn;
int gcd(int a,int b)
{
    return b == 0 ? a : gcd(b,a % b);
}

int a[maxn];
int num[M];
int p[4000],prime[4000];
void init()
{
    pn = 0;
    memset(p,0,sizeof p);
    for(int i=2;i<4000;i++)
    {
        if(!p[i])
            prime[pn++] = i;
        for(ll j =0 ;j < pn && i * prime[j] < 4000; j++){
            p[i * prime[j]] = 1;
            if(i % prime[j] == 0)
                continue;
        }
    }
    //cout<<pn<<endl;
}
int main()
{
    int n,mingcd;
    init();
    scanf("%d",&n);
    memset(num,0,sizeof num);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        if(i == 1)
            mingcd = a[i];
        else
            mingcd = gcd(mingcd,a[i]);
    }
    for(int i=1;i<=n;i++)
        a[i] /= mingcd;
    int ans = -1;
    for(int i=1;i<=n;i++)
    {
        for(int j=0;j<pn && prime[j] * prime[j] <= a[i];j++)
        {
            if(a[i] % prime[j] == 0)
            {
                num[prime[j]]++;
                ans = max(ans,num[prime[j]]);
                while(a[i] % prime[j] == 0)
                    a[i] /= prime[j];
            }
        }
        if(a[i] > 1)
        {
            num[a[i]]++;
            ans = max(ans,num[a[i]]);
        }
    }
    printf("%d
", ans == -1 ? ans : n - ans);
}

D - Little C Loves 3 II

 CodeForces - 1047D 

题意：给你n*m得棋盘，让你找两点之间距离为3的点的个数，不能重复使用，距离定义，两坐标差绝对值之和、

题解：找规律，找特殊样例即可

#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 = 1e3+7;

int gcd(int a,int b)
{
    return b == 0 ? a : gcd(b,a % b);
}
int main()
{
   ll n,m;
   ll ans;
   scanf("%lld %lld",&n,&m);
   if(n > m)
       swap(n,m);
   if(n == 1)
   {
       if(m % 6 == 0)
           ans = n * m;
       else if(m % 6 <= 3)
           ans = m - m % 6;
       else
           ans = m - (6 - m % 6);
   }
   else if(n == 2)
   {
       if(m == 2)
           ans = 0;
       else if(m == 3)
           ans = 4;
       else if(m == 7)
           ans = 12;
       else
           ans = n * m;
   }
   else
   {
       if(n * m % 2 == 1)
           ans = n * m - 1;
       else
           ans = n * m;
   }
   printf("%lld
",ans);
}

E - Region Separation

 CodeForces - 1047E 

题意：给定一棵大小为n的树，点有点权，在一个方案中可以将整棵树划分多次，要求每次划分后各个联通块的权值和相等，问有多少种划分方案

 

转自：https://blog.csdn.net/Mys_C_K/article/details/82867961

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

const int MOD = 1e9 + 7;

LL s[MAXN];
LL f[MAXN],p[MAXN],ans[MAXN];
LL gcd(LL a,LL b)
{
    return b == 0 ? a : gcd(b,a % b);
}
int main()
{
    int n;
    LL tot = 0;

    scanf("%d",&n);
    for(int i = 1; i <=n; i++)
        scanf("%lld",&s[i]);
    for(int i = 2; i <= n; i++)
        scanf("%lld",&p[i]);
    for(int i = n; i;i--)
        s[p[i]] += s[i];
    for(int i = n; i;i--)
        s[i] = s[1] / gcd(s[1],s[i]);
    for(int i = 1; i <= n; i++) {
        if (s[i] <= n) {
            f[s[i]]++;
        }
    }
    for(int i = n; i; i--)
        for(int j = 2 * i; j <= n; j += i){
            f[j] = (f[j] + f[i]) % MOD;
        }
    for(int i=ans[1]=1;i<=n;i++)
        if(f[i]>=i)
            for(int j=i*2;j<=n;j+=i)
                ans[j] = (ans[j] + ans[i]) % MOD;
    for(int i=1;i<=n;i++)
        if(f[i]>=i)
            tot = (tot + ans[i]) % MOD;
    printf("%lld
",tot);
}