HDU 6327&&18多校训练3I 【DP】

HDU 6327

Problem I. Random Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others) Total Submission(s): 56    Accepted Submission(s): 13   Problem Description There is a positive integer sequence a1,a2,...,an with some unknown positions, denoted by 0. Little Q will replace each 0 by a random integer within the range [1,m] equiprobably. After that, he will calculate the value of this sequence using the following formula :             Little Q is wondering what is the expected value of this sequence. Please write a program to calculate the expected value.     Input The first line of the input contains an integer T(1≤T≤10), denoting the number of test cases. In each test case, there are 2 integers n,m(4≤n≤100,1≤m≤100) in the first line, denoting the length of the sequence and the bound of each number. In the second line, there are n integers a1,a2,...,an(0≤ai≤m), denoting the sequence. In the third line, there are m integers v1,v2,...vm(1≤vi≤109), denoting the array v.     Output For each test case, print a single line containing an integer, denoting the expected value. If the answer is AB, please print C(0≤C<109+7) where A≡C×B(mod109+7).     Sample Input   2 6 8 4 8 8 4 6 5 10 20 30 40 50 60 70 80 4 3 0 0 0 0 3 2 4     Sample Output   8000 3     Source 2018 Multi-University Training Contest 3

题意：给定数组a，如果ai为0则小明从1到m中随机选择一个数。求的期望值。

题解：

 

#include
using namespace std;
typedef long long LL;
const LL MOD=1e9+7;
int p[110][110],a[110],l[110],r[110];
LL dp[2][101][101][101];
LL v[110];
vector di[110];
LL pod(LL x,LL n,LL MOD)
{
    LL ret=1;
    while(n)
    {
        if(n&1)ret=ret*x%MOD;
        n>>=1;
        x=x*x%MOD;
    }
    return ret;
}
int main()
{
    for(int i=1; i<=100; i++)
        for(int j=1; j<=100; j++)
            p[i][j]=__gcd(i,j);
    for(int i=1; i<=100; i++)
        for(int j=i; j<=100; j+=i)
            di[j].push_back(i);
    int TA;
    scanf("%d",&TA);
    while(TA--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            if(a[i]==0)
            {
                l[i]=1;
                r[i]=m;
            }
            else
            {
                l[i]=a[i];
                r[i]=a[i];
            }
        }

        for(int i=1; i<=m; i++)scanf("%lld",&v[i]);
        memset(dp,0,sizeof(dp));
        int w=0;
        for(int i=l[3]; i<=r[3]; i++)
        {
            for(int j=l[2]; j<=r[2]; j++)
            {
                for(int k=l[1]; k<=r[1]; k++)
                {
                    dp[w][i][p[i][j]][p[p[i][j]][k]]++;
                }
            }
        }
        for(int i=3; i