数学题
题目描述
背景
求 ∑ 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 0≤n≤1e9,1≤k≤1e5
题解
看到组合数以及
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\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=0∑n(ni)j=1∑kjiFi=j=1∑ki=0∑nCniFij−i=j=1∑ki=0∑nCni⋅55[(25+1)i+1−(21−5)i+1]j−i=55j=1∑kj−n[i=0∑nCni(25+1)i+1jn−i−i=0∑nCni(21−5)i+1jn−i]=55j=1∑kj−n[25+1i=0∑nCni(25+1)ijn−i−21−5i=0∑nCni(21−5)ijn−i]=55j=1∑kj−n[α(α+j)n−β(β+j)n](α=25+1,β=21−5)
而
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5为模
998244353
998244353
998244353的二次非剩余，不可以直接求解
解法一
因为
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F_i = {\sqrt{5} \over 5} [\alpha^{i + 1} - \beta^{i + 1}] \in Z^* (i + 1 \in Z^*)
Fi=55[αi+1−βi+1]∈Z∗(i+1∈Z∗)，所以
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\alpha^i - \beta^i = m \sqrt{5} (m \in Z^* , i \in Z^*)
αi−βi=m5(m∈Z∗,i∈Z∗)
因而
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\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)n−β(β+j)n=∑i=0nCni[αi+1−βi+1]jn−i=mj5(mj∈Z∗)
设
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(\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
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\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)n−β(β+j)n=25+1(xα+yα5)−21−5(xβ+yβ5)=21xα+25yα−21xβ+25yβ+(21xα+21yα+21xβ−21yβ)5
又由定义知
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x_\alpha = x_\beta , y_\alpha + y_\beta = 0
xα=xβ,yα+yβ=0，所以
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\alpha (\alpha + j)^n - \beta (\beta + j)^n = (x_\alpha + y_\alpha) \sqrt{5}
α(α+j)n−β(β+j)n=(xα+yα)5
因而
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{\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=1kj−n[α(α+j)n−β(β+j)n]=∑j=1kj−n(xα+yα)
令
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(\alpha + j)^i = x_i + y_i \sqrt{5}
(α+j)i=xi+yi5， 则
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(\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
即
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\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]
又
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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
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c_n = \alpha a^n + \beta b^n
cn=αan+βbn，则
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(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+1−abcn=(a+b)(αan+1+βbn+1)−ab(αan+βbn)=αan+2+βbn+2=cn+2
令
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c_{n , j} = {\sqrt{5} \over 5} j^{-n} [\alpha (\alpha + j)^n - \beta (\beta + j)^n]
cn,j=55j−n[α(α+j)n−β(β+j)n]，即
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a , b , \alpha , \beta
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{\alpha + j \over j} , {\beta + j \over j} , {\sqrt{5} \over 5} \alpha , - {\sqrt{5} \over 5} \beta
jα+j,jβ+j,55α,−55β
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a + b = 2 + {1 \over j} , ab = 1 + {1 \over j} - {1 \over j^2}
a+b=2+j1,ab=1+j1−j21
所以
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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+j1−j21)cn,j，即
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\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+j11−1−j1+j210][cn+1,jcn,j]
又
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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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