操作数 （数论 ：前n项的前n项和）

题目链接

解题报告：
这题思维难度较高。

#include
#define LL long long
#define pii pair
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int maxn=1e5+5;
const int inf=0x3f3f3f3f;
const LL MOD=1e9+7;

int n,k;
LL p[maxn],f[maxn];
LL qpow(LL x,LL p){
    LL ans=1;
    while(p){
        if(p&1) ans=ans*x%MOD;
        x=x*x%MOD;
        p>>=1;
    }
    return ans;
}

int main() {
    LL sum;
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++) scanf("%lld", &p[i]);
    for (int i = 1; i <= n; i++) f[i] = (0LL + k + i - 1) * qpow(i, MOD - 2) % MOD;
    for (int i = 1; i <= n; i++) {
        sum = p[i];
        for (int j = 1; j < i; j++) {
            p[j] = p[j] * f[i - j] % MOD;
            sum = (sum + p[j]) % MOD;
        }
        printf("%lld ", sum);
    }
    return 0;
}