51nod1229 序列求和 V2 数学

Posted 2020-12-03 mychael

题目链接

B51nod1229

题解

我们要求
[sumlimits_{i = 1}^{n}i^{k}r^{i}]
如果(r = 1)，就是自然数幂求和，上伯努利数即可(O(k^2))
否则，我们需要将式子进行变形
要与(n)无关
设
[F(k) = sumlimits_{i = 1}^{n} i^{k}r^{i}]
自然数幂应该是不用去动了，两边乘个(r)
[rF(k) = sumlimits_{i = 2}^{n + 1}r^{i}(i - 1)^{k}]
相减得
[ egin{aligned} (r - 1)F(k) &= r^{n + 1}n^{k} - r + sumlimits_{i = 2}^{n}r^{i}((i - 1)^{k} - i^{k}) &= r^{n + 1}n^{k} - r + sumlimits_{i = 2}^{n}r^{i}(sumlimits_{j = 0}^{k}{k choose j}(-1)^{k - j}i^{j} - i^{k}) &= r^{n + 1}n^{k} - r + sumlimits_{i = 2}^{n}r^{i}sumlimits_{j = 0}^{k - 1}{k choose j}(-1)^{k - j}i^{j} &= r^{n + 1}n^{k} - r + sumlimits_{i = 2}^{n}sumlimits_{j = 0}^{k - 1}{k choose j}(-1)^{k - j}i^{j}r^{i} &= r^{n + 1}n^{k} - r + sumlimits_{j = 0}^{k - 1}{k choose j}(-1)^{k - j}sumlimits_{i = 2}^{n}i^{j}r^{i} &= r^{n + 1}n^{k} - r + sumlimits_{j = 0}^{k - 1}{k choose j}(-1)^{k - j}(F(j) - r) \end{aligned} ]
故
[F(k) = frac{r^{n + 1}n^{k} - r + sumlimits_{j = 0}^{k - 1}{k choose j}(-1)^{k - j}(F(j) - r)}{r - 1}]
边界
[F(0) = sumlimits_{i = 1}^{n}r^{i} = rfrac{r^{n} - 1}{r -1}]
同样可以实现(O(k^2))递推

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<map>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define cls(s,v) memset(s,v,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
using namespace std;
const int maxn = 2010,maxm = 100005,INF = 0x3f3f3f3f;
inline LL read(){
    LL out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}
    return flag ? out : -out;
}
const int P = 1000000007;
LL F[maxn],B[maxn],fac[maxn],inv[maxn],fv[maxn],N = 2005;
inline LL qpow(LL a,LL b){
    LL re = 1; a %= P;
    for (; b; b >>= 1,a = 1ll * a * a % P)
        if (b & 1) re = 1ll * re * a % P;
    return re;
}
inline LL C(LL n,LL m){
    if (m > n) return 0;
    return 1ll * fac[n] * fv[m] % P * fv[n - m] % P;
}
void init(){
    fac[0] = fac[1] = fv[0] = fv[1] = inv[0] = inv[1] = 1;
    for (int i = 2; i <= N; i++){
        fac[i] = fac[i - 1] * i % P;
        inv[i] = 1ll * (P - P / i) * inv[P % i] % P;
        fv[i] = fv[i - 1] * inv[i] % P;
    }
    B[0] = 1;
    for (int k = 1; k < N; k++){
        for (int i = 0; i < k; i++)
            B[k] = (B[k] + C(k + 1,i) * B[i] % P) % P;
        B[k] = 1ll * (P - 1) * inv[k + 1] % P * B[k] % P;
    }
}
LL n,K,r;
void work1(){
    n %= P;
    LL tmp = n,ans = 0;
    for (int i = K; ~i; i--){
        ans = (ans + C(K + 1,i) * B[i] % P * tmp % P) % P;
        tmp = tmp * n % P;
    }
    ans = ans * inv[K + 1] % P;
    printf("%lld
",(ans + qpow(n,K)) % P);
}
void work2(){
    r %= P;
    LL tmp = qpow(r,n + 1),t,tt = 1,rv = qpow(r - 1,P - 2);
    F[0] = 1ll * (qpow(r,n) + P - 1) % P * rv % P * r % P;
    for (int k = 1; k <= K; k++){
        t = 0; tt = 1ll * tt * (n % P) % P;
        for (int j = 0; j < k; j++)
            t = (t + (((k - j) & 1) ? -1ll : 1ll) * C(k,j) * ((F[j] - r) % P) % P) % P;
        t = (t + P) % P;
        F[k] = ((tmp * tt % P - r) % P + t) % P * rv % P;
    }
    printf("%lld
",(F[K] + P) % P);
}
int main(){
    init();
    int T = read();
    while (T--){
        n = read(); K = read(); r = read();
        if (r == 1) work1();
        else work2();
    }
    return 0;
}