HDU 5628 Clarke and math dp+數學

Clarke and math

題目鏈接：

http://acm.hdu.edu.cn/showproblem.php?pid=5628

Description

Clarke is a patient with multiple personality disorder. One day, he turned into a mathematician, did a research on interesting things.
Suddenly he found a interesting formula. Given f(i),1≤i≤n, calculate
g(i)=∑i1∣i∑i2∣i1∑i3∣i2⋯∑ik∣ik−1f(ik) mod 1000000007(1≤i≤n)

Input

The first line contains an integer T(1≤T≤5), the number of test cases.
For each test case, the first line contains two integers n,k(1≤n,k≤100000).
The second line contains n integers, the ith integer denotes f(i),0≤f(i)<109+7.

Output

For each test case, print a line contained n integers, the ith integer represents g(i).

Sample Input

2
6 2
2 3 3 3 3 3
23 3
2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

Sample Output

2 7 7 15 7 23
2 9 9 24 9 39 9 50 24 39 9 102 9 39 39 90 9 102 9 102 39 39 9

Hint

題意

題解：

dp
dp[i][j]表示第i位置，選擇了j個不一樣的因子以後，可以得到的權值是多少
ans[i]=sigma C(k,j)*dp[i][j]
爲何呢？
咱們考慮傳遞了k次的sigma，實際上就是在枚舉因子，在這個數據範圍內，最多枚舉20個不一樣的因子，並且因子顯然是不斷遞減的（固然，這句話沒什麼用
而後腦補腦補，這個就是對的了……

官方題解確實看不懂……
弱智選手並不會xx卷積……

代碼

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+7;
const int mod = 1e9+7;
long long fac[maxn];
long long qpow(long long a,long long b)
{
    long long ans=1;a%=mod;
    for(long long i=b;i;i>>=1,a=a*a%mod)
        if(i&1)ans=ans*a%mod;
    return ans;
}
long long C(long long n,long long m)
{
    if(m>n||m<0)return 0;
    long long s1=fac[n],s2=fac[n-m]*fac[m]%mod;
    return s1*qpow(s2,mod-2)%mod;
}
int a[maxn];
int dp[maxn][22];
int K=20;
int main()
{
    fac[0]=1;
    for(int i=1;i<maxn;i++)
        fac[i]=fac[i-1]*i%mod;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        memset(dp,0,sizeof(dp));
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)
            dp[i][0]=a[i];
        for(int i=1;i<=n;i++)
            for(int j=i+i;j<=n;j+=i)
                for(int k=0;k<K;k++)
                    dp[j][k+1]=(dp[j][k+1]+dp[i][k])%mod;
        for(int i=1;i<=n;i++)
        {
            int ans = 0;
            for(int j=0;j<=K;j++)
                ans=(ans+1ll*C(m,j)*dp[i][j])%mod;
            if(i==n)printf("%d",ans);else printf("%d ",ans);
        }
        printf("\n");
    }
}