出处不详 数学题

数学题

题目描述

背景

∑ i = 0 n ( n i ) ∑ j = 1 k F i j i \sum_{i = 0}^n ( \begin{matrix} n \\ i \end{matrix} ) \sum_{j = 1}^k {F_i \over j^i} i=0n(ni)j=1kjiFi,其中 F i F_i Fi表示斐波那契数列的第 i i i项,且 F 0 = 1 , F 1 = 1 F_0 = 1, F_1 = 1 F0=1,F1=1 ( n i ) ( \begin{matrix} n \\ i \end{matrix} ) (ni)为从 n n n个不同元素中取出 i i i个元素的组合数

输入

一行,两个整数 n , k n , k n,k

输出

一个整数,表示答案(答案是一个有理数,请输出答案对质数 998244353 998244353 998244353取模的结果)

数据范围

0 ≤ n ≤ 1 e 9 , 1 ≤ k ≤ 1 e 5 0 \le n \le 1e9 , 1 \le k \le 1e5 0n1e9,1k1e5

题解

看到组合数以及 j − i j^{-i} ji很容易想到使用二项式定理化简要求的式子,即:
∑ i = 0 n ( n i ) ∑ j = 1 k F i j i = ∑ j = 1 k ∑ i = 0 n C n i F i j − i = ∑ j = 1 k ∑ i = 0 n C n i ⋅ 5 5 [ ( 5 + 1 2 ) i + 1 − ( 1 − 5 2 ) i + 1 ] j − i = 5 5 ∑ j = 1 k j − n [ ∑ i = 0 n C n i ( 5 + 1 2 ) i + 1 j n − i − ∑ i = 0 n C n i ( 1 − 5 2 ) i + 1 j n − i ] = 5 5 ∑ j = 1 k j − n [ 5 + 1 2 ∑ i = 0 n C n i ( 5 + 1 2 ) i j n − i − 1 − 5 2 ∑ i = 0 n C n i ( 1 − 5 2 ) i j n − i ] = 5 5 ∑ j = 1 k j − n [ α ( α + j ) n − β ( β + j ) n ] ( α = 5 + 1 2 , β = 1 − 5 2 ) \begin{aligned} \sum_{i = 0}^n ( \begin{matrix} n \\ i \end{matrix} ) \sum_{j = 1}^k {F_i \over j^i} & = \sum_{j = 1}^k \sum_{i = 0}^n C_n^i F_i j^{-i} \\ & = \sum_{j = 1}^k \sum_{i = 0}^n C_n^i \cdot {\sqrt{5} \over 5} [({\sqrt{5} + 1 \over 2})^{i + 1} - ({1 - \sqrt{5} \over 2})^{i + 1}] j^{-i} \\ & = {\sqrt{5} \over 5} \sum_{j = 1}^k j^{-n} [\sum_{i = 0}^n C_n^i ({\sqrt{5} + 1 \over 2})^{i + 1} j^{n - i} - \sum_{i = 0}^n C_n^i ({1 - \sqrt{5} \over 2})^{i + 1} j^{n - i}] \\ & = {\sqrt{5} \over 5} \sum_{j = 1}^k j^{-n} [{\sqrt{5} + 1 \over 2} \sum_{i = 0}^n C_n^i ({\sqrt{5} + 1 \over 2})^i j^{n - i} - {1 - \sqrt{5} \over 2} \sum_{i = 0}^n C_n^i ({1 - \sqrt{5} \over 2})^i j^{n - i}] \\ & = {\sqrt{5} \over 5} \sum_{j = 1}^k j^{-n} [\alpha (\alpha + j)^n - \beta (\beta + j)^n] (\alpha = {\sqrt{5} + 1 \over 2} , \beta = {1 - \sqrt{5} \over 2}) \end{aligned} i=0n(ni)j=1kjiFi=j=1ki=0nCniFiji=j=1ki=0nCni55 [(25 +1)i+1(215 )i+1]ji=55 j=1kjn[i=0nCni(25 +1)i+1jnii=0nCni(215 )i+1jni]=55 j=1kjn[25 +1i=0nCni(25 +1)ijni215 i=0nCni(215 )ijni]=55 j=1kjn[α(α+j)nβ(β+j)n](α=25 +1,β=215 )
5 5 5为模 998244353 998244353 998244353的二次非剩余,不可以直接求解

解法一

因为 F i = 5 5 [ α i + 1 − β i + 1 ] ∈ Z ∗ ( i + 1 ∈ Z ∗ ) F_i = {\sqrt{5} \over 5} [\alpha^{i + 1} - \beta^{i + 1}] \in Z^* (i + 1 \in Z^*) Fi=55 [αi+1βi+1]Z(i+1Z),所以 α i − β i = m 5 ( m ∈ Z ∗ , i ∈ Z ∗ ) \alpha^i - \beta^i = m \sqrt{5} (m \in Z^* , i \in Z^*) αiβi=m5 (mZ,iZ)
因而 α ( α + j ) n − β ( β + j ) n = ∑ i = 0 n C n i [ α i + 1 − β i + 1 ] j n − i = m j 5 ( m j ∈ Z ∗ ) \alpha (\alpha + j)^n - \beta (\beta + j)^n = \sum_{i = 0}^n C_n^i [\alpha^{i + 1} - \beta^{i + 1}] j^{n - i} = m_j \sqrt{5} (m_j \in Z^*) α(α+j)nβ(β+j)n=i=0nCni[αi+1βi+1]jni=mj5 (mjZ)
( α + j ) n = x α + y α 5 , ( β + j ) n = x β + y β 5 (\alpha + j)^n = x_\alpha + y_\alpha \sqrt{5} , (\beta + j)^n = x_\beta + y_\beta \sqrt{5} (α+j)n=xα+yα5 ,(β+j)n=xβ+yβ5 ,则 α ( α + j ) n − β ( β + j ) n = 5 + 1 2 ( x α + y α 5 ) − 1 − 5 2 ( x β + y β 5 ) = 1 2 x α + 5 2 y α − 1 2 x β + 5 2 y β + ( 1 2 x α + 1 2 y α + 1 2 x β − 1 2 y β ) 5 \alpha (\alpha + j)^n - \beta (\beta + j)^n = {\sqrt{5} + 1 \over 2} (x_\alpha + y_\alpha \sqrt{5}) - {1 - \sqrt{5} \over 2} (x_\beta + y_\beta \sqrt{5}) = {1 \over 2} x_\alpha + {5 \over 2} y_\alpha - {1 \over 2} x_\beta + {5 \over 2} y_\beta + ({1 \over 2} x_\alpha + {1 \over 2} y_\alpha + {1 \over 2} x_\beta - {1 \over 2} y_\beta) \sqrt{5} α(α+j)nβ(β+j)n=25 +1(xα+yα5 )215 (xβ+yβ5 )=21xα+25yα21xβ+25yβ+(21xα+21yα+21xβ21yβ)5
又由定义知 x α = x β , y α + y β = 0 x_\alpha = x_\beta , y_\alpha + y_\beta = 0 xα=xβ,yα+yβ=0,所以 α ( α + j ) n − β ( β + j ) n = ( x α + y α ) 5 \alpha (\alpha + j)^n - \beta (\beta + j)^n = (x_\alpha + y_\alpha) \sqrt{5} α(α+j)nβ(β+j)n=(xα+yα)5
因而 5 5 ∑ j = 1 k j − n [ α ( α + j ) n − β ( β + j ) n ] = ∑ j = 1 k j − n ( x α + y α ) {\sqrt{5} \over 5} \sum_{j = 1}^k j^{-n} [\alpha (\alpha + j)^n - \beta (\beta + j)^n] = \sum_{j = 1}^k j^{-n} (x_\alpha + y_\alpha) 55 j=1kjn[α(α+j)nβ(β+j)n]=j=1kjn(xα+yα)
( α + j ) i = x i + y i 5 (\alpha + j)^i = x_i + y_i \sqrt{5} (α+j)i=xi+yi5 , 则 ( α + j ) i + 1 = ( x i + y i 5 ) ( j + 5 + 1 2 ) = ( 1 2 + j ) x i + 5 2 y i + [ 1 2 x i + ( 1 2 + j ) y i ] 5 (\alpha + j)^{i + 1} = (x_i + y_i \sqrt{5})(j + {\sqrt{5} + 1 \over 2}) = ({1 \over 2} + j) x_i + {5 \over 2} y_i + [{1 \over 2}x_i + ({1 \over 2} + j) y_i] \sqrt{5} (α+j)i+1=(xi+yi5 )(j+25 +1)=(21+j)xi+25yi+[21xi+(21+j)yi]5
[ x i + 1 y i + 1 ] = [ 1 2 + j 5 2 1 2 1 2 + j ] [ x i y i ] \begin{bmatrix} x_{i + 1} \\ y_{i + 1} \end{bmatrix} = \begin{bmatrix} {1 \over 2} + j & {5 \over 2} \\ {1 \over 2} & {1 \over 2} + j \end{bmatrix} \begin{bmatrix} x_i \\ y_i \end{bmatrix} [xi+1yi+1]=[21+j212521+j][xiyi]
x 0 = 1 , y 0 = 0 x_0 = 1 , y_0 = 0 x0=1,y0=0,因而采用矩阵快速幂再求和即可
代码如下:

#include<cstdio>

#define il inline

typedef long long ll;

const int mod = 998244353;
const int X21 = 499122177;
const int X12 = 499122179;

int n, k, ans;

il int read() {
    int x = 0;
    char c = getchar();
    while(c < '0' || c > '9') c = getchar();
    while(c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return x;
}

il int qpow(int x, int y) {
    int ret = 1;
    for(; y; x = (ll)x * x % mod, y >>= 1)
        if(y % 2) ret = (ll)ret * x % mod;
    return ret;
}

il void Mul(int &x11, int &x12, int &x21, int &x22, int y11, int y12, int y21, int y22) {
    int r11 = ((ll)x11 * y11 % mod + (ll)x12 * y21 % mod) % mod;
    int r12 = ((ll)x11 * y12 % mod + (ll)x12 * y22 % mod) % mod;
    int r21 = ((ll)x21 * y11 % mod + (ll)x22 * y21 % mod) % mod;
    int r22 = ((ll)x21 * y12 % mod + (ll)x22 * y22 % mod) % mod;
    x11 = r11, x12 = r12, x21 = r21, x22 = r22;
}

il void qPow(int &x11, int x12, int &x21, int x22, int y) {
    if(y == 1) return ;
    int r11 = 1, r12 = 0, r21 = 0, r22 = 1;
    for(; y; Mul(x11, x12, x21, x22, x11, x12, x21, x22), y >>= 1)
        if(y % 2) Mul(r11, r12, r21, r22, x11, x12, x21, x22);
    x11 = r11, x21 = r21;
}

int main() {
    n = read() % mod, k = read();
    for(int i = 1; i <= k; ++i) {
        int x11 = X21 + i, x12 = X12, x21 = X21, x22 = x11;
        qPow(x11, x12, x21, x22, n);
        ans = (ans + (ll)(x11 + x21) * qpow(qpow(i, n), mod - 2) % mod) % mod;
    }
    printf("%d", ans);
    return 0;
}

解法二

c n = α a n + β b n c_n = \alpha a^n + \beta b^n cn=αan+βbn,则 ( a + b ) c n + 1 − a b c n = ( a + b ) ( α a n + 1 + β b n + 1 ) − a b ( α a n + β b n ) = α a n + 2 + β b n + 2 = c n + 2 (a + b) c_{n + 1} - ab c_n = (a + b)(\alpha a^{n + 1} + \beta b^{n + 1}) - ab(\alpha a^n + \beta b^n) = \alpha a^{n + 2} + \beta b^{n + 2} = c_{n + 2} (a+b)cn+1abcn=(a+b)(αan+1+βbn+1)ab(αan+βbn)=αan+2+βbn+2=cn+2
c n , j = 5 5 j − n [ α ( α + j ) n − β ( β + j ) n ] c_{n , j} = {\sqrt{5} \over 5} j^{-n} [\alpha (\alpha + j)^n - \beta (\beta + j)^n] cn,j=55 jn[α(α+j)nβ(β+j)n],即 a , b , α , β a , b , \alpha , \beta a,b,α,β分别为 α + j j , β + j j , 5 5 α , − 5 5 β {\alpha + j \over j} , {\beta + j \over j} , {\sqrt{5} \over 5} \alpha , - {\sqrt{5} \over 5} \beta jα+j,jβ+j,55 α,55 β
a + b = 2 + 1 j , a b = 1 + 1 j − 1 j 2 a + b = 2 + {1 \over j} , ab = 1 + {1 \over j} - {1 \over j^2} a+b=2+j1,ab=1+j1j21
所以 c n + 2 , j = ( 2 + 1 j ) c n + 1 , j − ( 1 + 1 j − 1 j 2 ) c n , j c_{n + 2 , j} = (2 + {1 \over j}) c_{n + 1 , j} - (1 + {1 \over j} - {1 \over j^2}) c_{n , j} cn+2,j=(2+j1)cn+1,j(1+j1j21)cn,j,即 [ c n + 2 , j c n + 1 , j ] = [ 2 + 1 j − 1 − 1 j + 1 j 2 1 0 ] [ c n + 1 , j c n , j ] \begin{bmatrix} c_{n + 2 , j} \\ c_{n + 1 , j} \end{bmatrix} = \begin{bmatrix} 2 + {1 \over j} & - 1 - {1 \over j} + {1 \over j^2} \\ 1 &0 \end{bmatrix} \begin{bmatrix} c_{n + 1 , j} \\ c_{n , j} \end{bmatrix} [cn+2,jcn+1,j]=[2+j111j1+j210][cn+1,jcn,j]
c 0 , j = 1 , c 1 , j = 1 + 1 j c_{0 , j} = 1 , c_{1 , j} = 1 + {1 \over j} c0,j=1,c1,j=1+j1,因而采用矩阵快速幂再求和即可
代码如下:

#include<cstdio>

#define il inline

typedef long long ll;

const int mod = 998244353;

int n, k, ans;

il int read() {
    int x = 0;
    char c = getchar();
    while(c < '0' || c > '9') c = getchar();
    while(c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return x;
}

il int qpow(int x, int y) {
    int ret = 1;
    for(; y; x = (ll)x * x % mod, y >>= 1)
        if(y % 2) ret = (ll)ret * x % mod;
    return ret;
}

il void Mul(int &x11, int &x12, int &x21, int &x22, int y11, int y12, int y21, int y22) {
    int r11 = ((ll)x11 * y11 % mod + (ll)x12 * y21 % mod) % mod;
    int r12 = ((ll)x11 * y12 % mod + (ll)x12 * y22 % mod) % mod;
    int r21 = ((ll)x21 * y11 % mod + (ll)x22 * y21 % mod) % mod;
    int r22 = ((ll)x21 * y12 % mod + (ll)x22 * y22 % mod) % mod;
    x11 = r11, x12 = r12, x21 = r21, x22 = r22;
}

il void qPow(int &x11, int &x12, int &x21, int &x22, int y) {
    if(y == 1) return ;
    int r11 = 1, r12 = 0, r21 = 0, r22 = 1;
    for(; y; Mul(x11, x12, x21, x22, x11, x12, x21, x22), y >>= 1)
        if(y % 2) Mul(r11, r12, r21, r22, x11, x12, x21, x22);
    x11 = r11, x12 = r12, x21 = r21, x22 = r22;
}

int main() {
    n = read() % mod, k = read();
    for(int i = 1; i <= k; ++i) {
        int D = qpow(i, mod - 2);
        int x11 = (2 + D) % mod, x12 = (-1 - D + mod + qpow(D, 2)) % mod, x21 = 1, x22 = 0;
        qPow(x11, x12, x21, x22, n - 1);
        Mul(x11, x12, x21, x22, 1 + D, 0, 1, 0);
        ans = (ans + x11) % mod;
    }
    printf("%d", ans);
    return 0;
}
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