2017 Multi-University Training Contest

2017 Multi-University Training Contest - Team 3

1011 RXD's date

 

Problem Description

 

As we all know that RXD is a life winner, therefore he always goes out, dating with his female friends.
Nevertheless, it is a pity that his female friends don't like extremely hot temperature. Due to this fact, they would not come out if it is higher than 35 degrees.
RXD is omni-potent, so he could precisely predict the temperature in the next t days, but he is poor in counting.
He wants to know, how many days are there in the next t days, in which he could go out and date with his female friends.

 

Input

There is only one test case.
The first line consists of one integer t.
The second line consists of t integers ci which means the temperature in the next t days.
1≤t≤1000 
0≤c_i≤50

 

Output

Output an integer which means the answer.

 

Sample Input

5

33 34 35 36 37

Sample Output

3

 

题意：让你求出在n个数中有多少个数小于等于35

 

题解：见过的多校最水的签到题之一，直接判断就可以了

 

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
    int t,i,x;
    while (scanf("%d",&t) != EOF)
    {
        int ans = 0;
        for (i = 0;i < t;i++)
        {
            scanf("%d",&x);
            if (x <= 35) ans++;    
        }    
        printf("%d
",ans);
    }    
} 

 

1008 RXD and math

Problem Description

RXD is a good mathematician.
One day he wants to calculate:

output the answer module 10⁹+7.
1 ≤ n, k ≤ 10¹⁸

PS：

μ(n) = 1 (n = 1)

μ(n) = (−1)^k (n = p₁p₂…p_k)

μ(n) = 0 (otherwise)
p₁,p₂,p₃…p_k are different prime numbers.

 

Input

There are several test cases, please keep reading until EOF.
There are exact 10000 cases.
For each test case, there are 2 numbers n,k.

 

Output

For each test case, output "Case #x: y", which means the test case number and the answer.

 

Sample Input

10 10

 

Sample Output

Case #1: 999999937

 

题意：给出μ(i)的定义为莫比乌斯函数，请计算这个式子。

 

题解：通过打表可知，这个式子其实是有规律的，结果即为n^k，因此直接裸快速幂即可解决。

 

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
long long n,k;
const long long mod = 1e9 + 7;
long long quick_power(long long a,long long b)  
{  
    long long res = 1;  
    while (b > 0)  
    {  
        if (b & 1) res = (res * a) % mod;  
        a = (a * a) % mod;  
        b = b >> 1;  
    }  
    return res;  
}  
int main()
{
    int count = 0;
    while (scanf("%lld%lld",&n,&k) != EOF)
    {
        count++;
        n = n % mod;
        printf("Case #%d: %lld
",count,quick_power(n,k));
    }
}

 

 1005 RXD and dividing

Problem Description

RXD has a tree T, with the size of n. Each edge has a cost.
Define f(S) as the the cost of the minimal Steiner Tree of the set S on tree T. 
he wants to divide 2,3,4,5,6,…n into k parts S₁,S₂,S₃,…S_k,
where ⋃S_i = {2,3,…,n} and for all different i,j , we can conclude that S_i ⋂ S_j = ∅. 
Then he calulates res = ∑f({1} ⋃ S_i).
He wants to maximize the res.
1 ≤ k ≤ n ≤ 106
the cost of each edge ∈ [1,105]
S_i might be empty.
f(S) means that you need to choose a couple of edges on the tree to make all the points in S connected, and you need to minimize the sum of the cost of these edges. f(S) is equal to the minimal cost.

 

Input

There are several test cases, please keep reading until EOF.
For each test case, the first line consists of 2 integer n,k, which means the number of the tree nodes , and k means the number of parts.
The next n−1 lines consists of 2 integers, a,b,c, means a tree edge (a,b) with cost c.
It is guaranteed that the edges would form a tree.
There are 4 big test cases and 50 small test cases.
small test case means n≤100.

 

Output

For each test case, output an integer, which means the answer.

 

Sample Input

5 4

1 2 3

2 3 4

2 4 5

2 5 6

Sample Output

27

 

题意：给出一棵树，有n个点n - 1条边，设f(S)为树T的最小斯坦纳树的边权和。将2, 3, ..., n分成k个两两不相交的集合S₁, S₂, ..., S_k， 将1插入这些集合，求出这些集合的最小生成树，并使其权值和最大，求出这个值。

 

补充：

       斯坦纳树：

       斯坦纳树问题是组合优化学科中的一个问题。将指定点集合中的所有点连通，且边权总和最小的生成树称为最小斯坦纳树（Minimal Steiner Tree），其实最小生成树是最小斯坦纳树的一种特殊情况。而斯坦纳树可以理解为使得指定集合中的点连通的树，但不一定最小。（最小生成树是在给定的点集和边中寻求最短网络使所有点连通。而最小斯坦纳树允许在给定点外增加额外的点，使生成的最短网络开销最小。）

 

题解：以1为根建树，找出每条边下面有多少个点，若这条边下面的点数num > k， 则这条边需要被算k次；否则取num个点。这就相当于将这些点放到不同的集合中去，然后这些边可以多算几次。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<vector>
using namespace std;
int n,k;
vector <int> map[1000050];
vector <int> dist[1000050];
int num[1000050];
long long ans[1000050];
void dfs(int x,int fa,int cost)
{
    int i;
    num[x] = 1;
    for (i = 0;i < map[x].size();i++)
    {
        if (map[x][i] != fa)
        {
            dfs(map[x][i],x,dist[x][i]);
            ans[x] = ans[x] + ans[map[x][i]];
            num[x] = num[x] + num[map[x][i]]; 
        }
    }
    if (num[x] >= k) ans[x] = ans[x] + 1LL * k * cost;
    else ans[x] = ans[x] + 1LL * num[x] * cost;
}
int main()
{
    int i,x,y,z;
    while (scanf("%d%d",&n,&k) != EOF)
    {
        for (i = 0;i <= n;i++) map[i].clear();
        for (i = 0;i <= n;i++) dist[i].clear();
        memset(num,0,sizeof(num));
        memset(ans,0,sizeof(ans));
        for (i = 1;i <= n - 1;i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            map[x].push_back(y);
            dist[x].push_back(z);
            map[y].push_back(x);
            dist[y].push_back(z);
        }
        dfs(1,-1,0);
        printf("%lld
",ans[1]);
    }
}