Clarke and math
题目连接：
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
题意
题解：
dp
dp[i][j]表示第i位置，选择了j个不同的因子之后，能够获得的权值是多少
ans[i]=sigma C(k,j)*dp[i][j]
为什么呢？
我们考虑传递了k次的sigma，实际上就是在枚举因子，在这个数据范围内，最多枚举20个不同的因子，而且因子显然是不断递减的（当然，这句话没什么用
然后脑补脑补，这个就是对的了……
官方题解确实看不懂……
弱智选手并不会xx卷积……
代码
#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("
");
}
}