ZOJ Monthly, March 2018

Easy Number Game

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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The bored BaoBao is playing a number game. In the beginning, there are  numbers. For each turn, BaoBao will take out two numbers from the remaining numbers, and calculate the product of them.

Now, BaoBao is curious to know the minimum sum of the products if he plays at least  turns. Can you tell him?

Input

The first line of input contains a positive integer  (about 30), the number of test cases. For each test case:

The first line contains two integers  and  (, ). Their meanings are described above.

The second line contains  integers  (), indicating the numbers.

Output

For each test case output one integer, indicating the minimum sum of the products.

Sample Input

3
4 2
1 3 2 4
3 1
2 3 1
4 0
1 3 2 4

Sample Output

10
2
0

Hint

For the first sample test case, the answer is 1 × 4 + 3 × 2 = 10.

For the second sample test case, the answer is 2 × 1 = 2.

n个数找m组数，使乘积的和最小

让前m小最大和最小的相乘就好了

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int a[N];
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int n,m;
        cin>>n>>m;
        for(int i=0; i<n; i++)
            cin>>a[i];
        sort(a,a+n);
        int r=2*m-1;
        long long s=0;
        for(int i=0; i<m; i++)
            s+=a[i]*a[r-i];
        cout<<s<<endl;
    }
}

Super Brain

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Time Limit: 1 Second      Memory Limit: 65536 KB

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Super Brain is a famous scientific reality and talent show aiming to find people with exceptional brainpower.

In one of the challenges, two integer sequences  and  of length  are given to the contestant. It's guaranteed that  and  hold for all , and there is exactly one integer in the first sequence which also appears in the second sequence. The contestant has to memorize the two sequences in a very short time, and find the integer which appears in both sequences correctly.

As a technical staff of the show, you are required to write a program and find out the correct integer.

Input

There are multiple test cases. The first line of the input contains an integer , indicating the number of test cases. For each test case:

The first line contains an integer  (), indicating the length of the sequence.

The second line contains  integers  (), indicating the first sequence.

The third line contains  integers  (), indicating the second sequence.

It's guaranteed that  and  hold for all , and there is exactly one integer in the first sequence which also appears in the second sequence.

It's also guaranteed that the sum of  over all test cases will not exceed .

Output

For each test case output one line containing one integer, indicating the integer which appears in both sequences.

Sample Input

3
3
3 1 2
5 3 4
2
38324 14122
38323 14122
1
180310
180310

Sample Output

3
14122
180310

我可能被卡常了，memset就过了
找到两个数组唯一相同的

#include<bits/stdc++.h>
using namespace std;
const int N=1e6+5;
int a[N],n;
int main()
{
    ios::sync_with_stdio(false);
    int T;
    cin>>T;
    while(T--)
    {
        memset(a,0,sizeof a);
        cin>>n;
        int ans;
        for(int i=0,x; i<n; i++)cin>>x,a[x]++;
        for(int i=0,x; i<n; i++)
        {
            cin>>x;
            if(a[x])ans=x;
        }
        cout<<ans<<"\n";
    }
}

Happy Sequence

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Time Limit: 3 Seconds      Memory Limit: 65536 KB

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A sequence of  integers  () is called a happy sequence if each number divides (without a remainder) the next number in the sequence. More formally, we can say  for all , or we can say  for all .

Given  and , find the number of happy sequences of length . Two sequences  and  are different, if and only if there exists an such that  and .

As the answer can be rather large print it modulo  ().

Input

There are multiple test cases. The first line of the input contains an integer  (about 50), indicating the number of test cases. For each test case:

The first and only line contains two integers  and  (), indicating the upper limit of the elements in the sequence and the length of the sequence.

Output

For each case output a single integer, indicating the number of happy sequences of length  modulo 

Sample Input

1
3 2

Sample Output

5

Hint

In the sample test case, the happy sequences are: , , , , .

 一个数组是按照倍数关系的

这个题利用埃筛的思想去枚举i个数s*j作为最后一个元素就好

using namespace std;
typedef long long ll;
const ll MD=1e9+7;
const int N=2005;
ll dp[N][N];
int main()
{
    ios::sync_with_stdio(false);
    int T;
    cin>>T;
    while(T--)
    {
        int n,k;
        cin>>n>>k;
        memset(dp,0,sizeof dp);
        for(int i=1; i<N; i++) dp[1][i]=1;
        for(int i=1; i<k; i++)
            for(int j=1; j<=n; j++)
                for(int s=1; s*j<=n; s++)
                    dp[i+1][s*j]=(dp[i+1][s*j]+dp[i][j])%MD;
        ll sum=0;
        for(int i=1; i<=n; i++) sum=(sum+dp[k][i])%MD;
        cout<<(sum+MD)%MD<<endl;
    }
    return 0;
}

Travel along the Line

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Time Limit: 1 Second      Memory Limit: 65536 KB

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BaoBao is traveling along a line with infinite length.

At the beginning of his trip, he is standing at position 0. At the beginning of each second, if he is standing at position , with  probability he will move to position , with  probability he will move to position , and with  probability he will stay at position . Positions can be positive, 0, or negative.

DreamGrid, BaoBao's best friend, is waiting for him at position . BaoBao would like to meet DreamGrid at position  after exactly  seconds. Please help BaoBao calculate the probability he can get to position  after exactly  seconds.

It's easy to show that the answer can be represented as , where  and  are coprime integers, and  is not divisible by . Please print the value of  modulo , where  is the multiplicative inverse of  modulo .

Input

There are multiple test cases. The first line of the input contains an integer  (about 10), indicating the number of test cases. For each test case:

The first and only line contains two integers  and  (). Their meanings are described above.

Output

For each test case output one integer, indicating the answer.

Sample Input

3
2 -2
0 0
0 1

Sample Output

562500004
1
0

---

1s分为2s，只要n+m挑对了就行了

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MD=1e9+7;
const int N=2e5+5;
ll la(ll a,ll b)
{
    a%=MD;
    ll ans=1;
    while(b>0)
    {
        if(b&1)ans=ans*a%MD;
        b>>=1;
        a=a*a%MD;
    }
    return ans;
}
ll f[N],v[N],ans;
ll C(ll n,ll m)
{
    if(m<0||m>n) return 0;
    return f[n]*v[m]%MD*v[n-m]%MD;
}
int main()
{
    ios::sync_with_stdio(false);
    f[0]=1;
    for (ll i=1; i<N; i++) f[i]=f[i-1]*i%MD;
    v[N-1]=la(f[N-1],MD-2);
    for (ll i=N-2; i>=0; i--) v[i]=v[i+1]*(i+1LL)%MD;
    int T;
    cin>>T;
    while(T--)
    {
        int n,m;
        cin>>n>>m;
        ans=C(2*n,m+n)*la(la(2,2*n),MD-2)%MD;
        cout<<ans<<"\n";
    }
    return 0;
}

 

Lucky Man

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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BaoBao is keen on collection. Recently he is abandoning himself to Kantai Collection, especially to collecting cute girls, known as "Fleet Girls".

There are  various types of girls in the game. To get a girl, one can use some materials to build her. The probability to get a type of girl by building is the same for all types of girls. From the Coupon Collector's Problem we know that, to collect all types of girls, the expected number of times of building is .

But this rule does not apply to BaoBao, as he is always luckier than the ordinary players (maybe because he's an European). For BaoBao to collect all types of girls, the expected number of times of building is , where  means the maximum integer that doesn't exceed .

As a lucky man, BaoBao is not interested in the actual value of , and he just wants to know whether  is odd or even. Can you help him?

Input

The first line of the input is an interger  (about 100), indicating the number of test cases. For each test case:

The first line contains an integer  (), indicating the number of types of girls.

Output

For each test case, if  is odd output "1" (without quotes), if  is even output "0" (without quotes).

Sample Input

9
2
3
23
233
2333
23333
233333
2333333
23333333

Sample Output

1
1
0
1
0
0
1
1
0

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被卡了一波时间，开方啊

def sqrt(n):
    l=0
    r=n
    while l<=r :
        mi=(l+r)>> 1
        F=mi*mi
        if(F<n):
            l=mi+1
        elif(F>n): r=mi-1
        else:
            return mi
    return l-1
T=int(input())
for i in range(0,T):
    a=int(input())
    if a==0:
        print("0")
    elif a<=3:
        print("1")
    else:
        a=sqrt(a)
        print(a & 1)

 sqrt对于每个都是成立的

def sqrt(n):
    l=0
    r=n
    while l<=r :
        mi=(l+r)>> 1
        F=mi*mi
        if(F<n):
            l=mi+1
        elif(F>n): r=mi-1
        else:
            return mi
    return l-1
T=int(input())
for i in range(0,T):
    a=int(input())
    a=sqrt(a)
    print(a & 1)