\(\color{#0066ff}{ 题目描述 }\)
一个\(n\)位数，每位可以是给出的\(k\)个数码中的一个数，可以有前导\(0\),输出前\(n/2\)位之和与后\(n/2\)位之和相等的方案数，保证\(n\)是偶数。
\(\color{#0066ff}{输入格式}\)
输入的第一行是两个整数\(n,k\)
接下来的一行有\(k\)个数\(d_1,d_2,\cdots,d_k(0\leq d_i\leq 9)\)
\(\color{#0066ff}{输出格式}\)
输出一个数，为方案数模\(998244353\)的值
\(\color{#0066ff}{输入样例}\)
4 2
1 8
20 1
6
10 5
6 1 4 0 3
1000 7
5 4 0 1 8 3 2
\(\color{#0066ff}{输出样例}\)
6
1
569725
460571165
\(\color{#0066ff}{数据范围与提示}\)
\(2\leq n\leq 2*10^5,1\leq k\leq 10\)
\(\color{#0066ff}{ 题解 }\)
构造生成函数
比如第一个样例，构造出\(x^1+x^8\)
然后序列的一半是2，就让它平方是\(x^2+2*x^9+x^{16}\)
每个x指数是凑出的和，前面的系数就是方案数，但是这是一半的，乘法原理
\(ans=\sum 系数平方\)
直接暴力快速幂套FFT可过
#include<bits/stdc++.h>
#define LL long long
LL in() {
char ch; LL x = 0, f = 1;
while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
return x * f;
}
using std::vector;
const int mod = 998244353;
const int maxn = 8e6 + 10;
LL ksm(LL x, LL y) {
LL re = 1LL;
while(y) {
if(y & 1) re = re * x % mod;
x = x * x % mod;
y >>= 1;
}
return re;
}
int r[maxn], len;
void FNTT(vector<int> &A, int flag) {
A.resize(len);
for(int i = 0; i < len; i++) if(i < r[i]) std::swap(A[i], A[r[i]]);
for(int l = 1; l < len; l <<= 1) {
int w0 = ksm(3, (mod - 1) / (l << 1));
for(int i = 0; i < len; i += (l << 1)) {
int w = 1, a0 = i, a1 = i + l;
for(int k = 0; k < l; k++, a0++, a1++, w = 1LL * w * w0 % mod) {
int tmp = 1LL * A[a1] * w % mod;
A[a1] = ((A[a0] - tmp) % mod + mod) % mod;
A[a0] = (A[a0] + tmp) % mod;
}
}
}
if(flag == -1) {
std::reverse(A.begin() + 1, A.end());
int inv = ksm(len, mod - 2);
for(int i = 0; i < len; i++) A[i] = 1LL * A[i] * inv % mod;
}
}
vector<int> operator * (vector<int> A, vector<int> B) {
int tot = A.size() + B.size() - 1;
for(len = 1; len <= tot; len <<= 1);
for(int i = 0; i < len; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) * (len >> 1));
FNTT(A, 1), FNTT(B, 1);
vector<int> ans;
ans.resize(len);
for(int i = 0; i < len; i++) ans[i] = 1LL * A[i] * B[i] % mod;
FNTT(ans, -1);
ans.resize(tot);
return ans;
}
vector<int> ksm(vector<int> A, int B) {
vector<int> ans;
ans.push_back(1);
while(B) {
if(B & 1) ans = ans * A;
A = A * A;
B >>= 1;
}
return ans;
}
int main() {
int n = in(), k = in();
vector<int> ans;
ans.resize(10);
for(int i = 1; i <= k; i++) ans[in()] = 1;
ans = ksm(ans, n >> 1);
LL tot = 0;
for(int i = 0; i < (int)ans.size(); i++) (tot += 1LL * ans[i] * ans[i] % mod) %= mod;
printf("%lld\n", tot);
return 0;
}
----olinr
