C. Palindrome Basis

题意

定义一个正整数 $a$ 是回文的（没有前导 $0$）当且仅当：$a$ 的十进制表示形式回文

给定一个正整数 $n$ ，求出将 $n$ 拆分成若干个回文数之和的方案数

思路

这是一个经典模型，与爬楼梯问题不同的是：这道题一个物品的选择先后顺序无关
在 $n\le 4\cdot 10^4$ 时，求出回文数字共有 $498$ 个，考虑 $DP$：
用 $dp[i][j]$ 表示只使用前 $j$ 个回文数字来构成 $i$ 的方案数，那么转移为：

$dp[i][j]=dp[i][j-1]+dp[i-P[j][j]，P[j]\le i$

意思就是：不用第 $j$ 个回文数字 加上 使用 $P[j]$ 的方案数。

初始化：$\forall j\in [1,498],dp[0][j]=1$

#include<bits/stdc++.h>
#define fore(i,l,r)	for(int i=(int)(l);i<(int)(r);++i)
#define fi first
#define se second
#define endl '\n'
#define ull unsigned long long
#define ALL(v) v.begin(), v.end()
#define Debug(x, ed) std::cerr << #x << " = " << x << ed;const int INF=0x3f3f3f3f;
const long long INFLL=1e18;typedef long long ll;template<class T>
constexpr T power(T a, ll b){T res = 1;while(b){if(b&1) res = res * a;a = a * a;b >>= 1;}return res;
}constexpr ll mul(ll a,ll b,ll mod){ //快速乘，避免两个long long相乘取模溢出ll res = a * b - ll(1.L * a * b / mod) * mod;res %= mod;if(res < 0) res += mod; //误差return res;
}template<ll P>
struct MLL{ll x;constexpr MLL() = default;constexpr MLL(ll x) : x(norm(x % getMod())) {}static ll Mod;constexpr static ll getMod(){if(P > 0) return P;return Mod;}constexpr static void setMod(int _Mod){Mod = _Mod;}constexpr ll norm(ll x) const{if(x < 0){x += getMod();}if(x >= getMod()){x -= getMod();}return x;}constexpr ll val() const{return x;}explicit constexpr operator ll() const{ return x; //将结构体显示转换为ll类型： ll res = static_cast<ll>(OBJ)}constexpr MLL operator -() const{ //负号，等价于加上ModMLL res;res.x = norm(getMod() - x);return res;}constexpr MLL inv() const{assert(x != 0);return power(*this, getMod() - 2); //用费马小定理求逆}constexpr MLL& operator *= (MLL rhs) & { //& 表示“this”指针不能指向一个临时对象或const对象x = mul(x, rhs.x, getMod()); //该函数只能被一个左值调用return *this;}constexpr MLL& operator += (MLL rhs) & {x = norm(x + rhs.x);return *this;}constexpr MLL& operator -= (MLL rhs) & {x = norm(x - rhs.x);return *this;}constexpr MLL& operator /= (MLL rhs) & {return *this *= rhs.inv();}friend constexpr MLL operator * (MLL lhs, MLL rhs){MLL res = lhs;res *= rhs;return res;}friend constexpr MLL operator + (MLL lhs, MLL rhs){MLL res = lhs;res += rhs;return res;}friend constexpr MLL operator - (MLL lhs, MLL rhs){MLL res = lhs;res -= rhs;return res;}friend constexpr MLL operator / (MLL lhs, MLL rhs){MLL res = lhs;res /= rhs;return res;}friend constexpr std::istream& operator >> (std::istream& is, MLL& a){ll v;is >> v;a = MLL(v);return is;}friend constexpr std::ostream& operator << (std::ostream& os, MLL& a){return os << a.val();}friend constexpr bool operator == (MLL lhs, MLL rhs){return lhs.val() == rhs.val();}friend constexpr bool operator != (MLL lhs, MLL rhs){return lhs.val() != rhs.val();}
};const ll mod = 1e9 + 7;
using Z = MLL<mod>;bool f(int x){if(x < 10) return true;std::string s = std::to_string(x);int len = s.size();fore(i, 0, len / 2)if(s[i] != s[len - 1 - i])return false;return true;
}int main(){std::ios::sync_with_stdio(false);std::cin.tie(nullptr);std::cout.tie(nullptr);std::vector<int> PN;PN.push_back(0);fore(i, 1, 40001)if(f(i))PN.push_back(i);std::vector<std::vector<Z>> dp(40005, std::vector<Z>(505, 0));fore(j, 1, 500) dp[0][j] = 1;fore(i, 1, 40001)fore(j, 1, 500)if(PN[j] <= i) dp[i][j] = dp[i][j - 1] + dp[i - PN[j]][j];else dp[i][j] = dp[i][j - 1];int t;std::cin >> t;while(t--){int n;std::cin >> n;std::cout << dp[n][498] << endl;}return 0;
}