\(\color{#0066ff}{ 题目描述 }\)

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 21 820 1610 56 1 4 0 31000 75 4 0 1 8 3 2

\(\color{#0066ff}{输出样例}\)

61   569725460571165

\(\color{#0066ff}{数据范围与提示}\)

\(2\leq n\leq 2*10^5,1\leq k\leq 10\)

\(\color{#0066ff}{ 题解 }\)

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
    
     #define LL long longLL 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
     
       &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
      
        operator * (vector
       
         A, vector
        
          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
         
           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
          
            ksm(vector
           
             A, int B) { vector
            
              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
             
               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;}