2014 Asia AnShan Regional Contest （dp）

Hatsune Miku

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 705    Accepted Submission(s): 506

Problem Description

Hatsune Miku is a popular virtual singer. It is very popular in both Japan and China. Basically it is a computer software that allows you to compose a song on your own using the vocal package.

Today you want to compose a song, which is just a sequence of notes. There are only m different notes provided in the package. And you want to make a song with n notes.


Also, you know that there is a system to evaluate the beautifulness of a song. For each two consecutive notes a and b, if b comes after a, then the beautifulness for these two notes is evaluated as score(a, b).

So the total beautifulness for a song consisting of notes a₁, a₂, . . . , a_n, is simply the sum of score(a_i, a_i+1) for 1 ≤ i ≤ n - 1.

Now, you find that at some positions, the notes have to be some specific ones, but at other positions you can decide what notes to use. You want to maximize your song’s beautifulness. What is the maximum beautifulness you can achieve?

 

Input

The first line contains an integer T (T ≤ 10), denoting the number of the test cases.

For each test case, the first line contains two integers n(1 ≤ n ≤ 100) and m(1 ≤ m ≤ 50) as mentioned above. Then m lines follow, each of them consisting of m space-separated integers, the j-th integer in the i-th line for score(i, j)( 0 ≤ score(i, j) ≤ 100). The next line contains n integers, a₁, a₂, . . . , a_n (-1 ≤ a_i ≤ m, a_i ≠ 0), where positive integers stand for the notes you cannot change, while negative integers are what you can replace with arbitrary notes. The notes are named from 1 to m.

 

Output

For each test case, output the answer in one line.

 

Sample Input

25 383 86 7715 93 3586 92 493 3 3 1 210 536 11 68 67 2982 30 62 23 6735 29 2 22 5869 67 93 56 1142 29 73 21 19-1 -1 5 -1 4 -1 -1 -1 4 -1

 

Sample Output

270625

题意：n个片段，组成一段音乐，音乐的score是score(ai,a[i+1])的和，现在告诉你矩阵wi,j表示ij组合的分数，下面n个片段，如果为负数，你可以自己安排

为最大的score是多少

思路：dp[i][j]表示前i个，最后一个是j的最大得分

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;const int maxn=110;int g[maxn][maxn],a[maxn];int N,M;int dp[maxn][55];int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&N,&M);        for(int i=1;i<=M;i++)            for(int j=1;j<=M;j++)scanf("%d",&g[i][j]);        for(int i=1;i<=N;i++)scanf("%d",&a[i]);        memset(dp,0,sizeof(dp));        for(int i=2;i<=N;i++)        {            if(a[i]>0)            {                if(a[i-1]>0)dp[i][a[i]]=max(dp[i][a[i]],dp[i-1][a[i-1]]+g[a[i-1]][a[i]]);                else                {                    for(int j=1;j<=M;j++)                        dp[i][a[i]]=max(dp[i][a[i]],dp[i-1][j]+g[j][a[i]]);                }            }            else            {                if(a[i-1]>0)                {                    for(int j=1;j<=M;j++)                        dp[i][j]=max(dp[i][j],dp[i-1][a[i-1]]+g[a[i-1]][j]);                }                else                {                    for(int j=1;j<=M;j++)                    {                        for(int k=1;k<=M;k++)                            dp[i][j]=max(dp[i][j],dp[i-1][k]+g[k][j]);                    }                }            }        }        int ans=0;        if(a[N]>0)printf("%d\n",dp[N][a[N]]);        else        {            for(int i=1;i<=M;i++)                ans=max(ans,dp[N][i]);            printf("%d\n",ans);        }    }    return 0;}

Coprime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1103    Accepted Submission(s): 447

Problem Description

There are n people standing in a line. Each of them has a unique id number.

Now the Ragnarok is coming. We should choose 3 people to defend the evil. As a group, the 3 people should be able to communicate. They are able to communicate if and only if their id numbers are pairwise coprime or pairwise not coprime. In other words, if their id numbers are a, b, c, then they can communicate if and only if [(a, b) = (b, c) = (a, c) = 1] or [(a, b) ≠ 1 and (a, c) ≠ 1 and (b, c) ≠ 1], where (x, y) denotes the greatest common divisor of x and y.

We want to know how many 3-people-groups can be chosen from the n people.

 

Input

The first line contains an integer T (T ≤ 5), denoting the number of the test cases.

For each test case, the first line contains an integer n(3 ≤ n ≤ 10⁵), denoting the number of people. The next line contains n distinct integers a₁, a₂, . . . , a_n(1 ≤ a_i ≤ 10⁵) separated by a single space, where a_i stands for the id number of the i-th person.

 

Output

For each test case, output the answer in a line.

 

Sample Input

151 3 9 10 2

 

Sample Output

4

题意：从N个数选三个，使他们两两互素或者两两不互素，有多少种选法

思路：设这三个数为a,b,c，现在假设a与b互质，b与c不互质，设满足这样的集合数位temp,那么答案为C(n,3)-temp/2，之所以除2是因为b作为中间的数算重了一次。 现在问题转为求解temp。  对于每个b，设与它互质的数个数为cnt1,与它不互质的个数为cnt2，显然cnt1+cnt2+1=n。 那么对于b的temp值为cnt1*cnt2，其中只需要求出cnt2即可，将b分解质因子，然后这些质因子的个数不会超过6个，显然可以状压以后容斥原理求解。 现在问题进而转化为求出一个数x有多少个数是x的倍数，这个用素数筛选可以求得，问题得以解决。

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;const int maxn=100010;typedef long long LL;int N,a[maxn],num;int prime[maxn];bool is_prime[maxn];int have[maxn];void getprime(){    num=0;    memset(is_prime,0,sizeof(is_prime));    is_prime[1]=1;    for(int i=2;i<maxn;i++)    {        if(!is_prime[i])            prime[num++]=i;        for(int j=0;j<num;j++)        {            if(i*prime[j]>maxn)break;            is_prime[i*prime[j]]=1;            if(i%prime[j]==0)break;        }    }}LL cal(int x){    vector<int>  q;    for(int i=0;i<num&&x>=prime[i];i++)    {        if(x%prime[i]==0)q.push_back(prime[i]);        while(x%prime[i]==0)x/=prime[i];        if(!is_prime[x]){q.push_back(x);break;}    }    int len=q.size();    LL ans=0;    for(int s=1;s<(1<<len);s++)    {        int cnt=0;        int u=1;        for(int i=0;i<len;i++)            if(s&(1<<i))                cnt++,u*=q[i];        if(cnt&1)ans=ans+have[u];        else ans=ans-have[u];    }    if(ans)ans--;    return ans*(N-1-ans);}int main(){    int T;    scanf("%d",&T);    getprime();    while(T--)    {        scanf("%d",&N);        for(int i=1;i<=N;i++)scanf("%d",&a[i]);        memset(have,0,sizeof(have));        for(int i=1;i<=N;i++)        {            int x=a[i];            for(int j=1;j*j<=a[i];j++)                if(a[i]%j==0)                {                    have[j]++;                    if(a[i]/j!=j)have[a[i]/j]++;                }        }        LL ans=0;        for(int i=1;i<=N;i++)            ans+=cal(a[i]);        printf("%I64d\n",(LL)N*(N-1)*(N-2)/6-ans/2);    }    return 0;}