传送门
题目描述
求出满足长度为 N N N,和为 M M M,并且异或为 0 0 0的序列个数。
分析
按照 2 k 2^k 2k拆开,因为异或为 0 0 0,所以每一组 2 k 2^k 2k只能选择偶数个,分组背包跑一下,预处理一下组合数就可以了
代码
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 5010;
const ll mod = 998244353;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
ll ksm (ll a , ll b) {
ll ans = 1 , base = a;
while (b) {if (b & 1) ans = ans * base % mod; b >>= 1; base = base * base % mod;} return ans;
}
ll f[N], inv[N];
ll dp[N][N];
int n, m;
ll C (ll a, ll b) {
if (a < b) return 0;
return f[a] * inv[a - b] % mod * inv[b] % mod;
}
int main() {
f[0] = inv[0] = 1;
for (int i = 1 ; i < N ; i++) {
f[i] = f[i - 1] * i % mod;
inv[i] = ksm(f[i] , mod - 2);
}
read(n), read(m);
for (int i = 0; i <= min(n, m); i += 2)
dp[0][i] = C(n, i);
for (int i = 1; i <= 22; i++)
for (int j = 0; j <= m; j++)
for (int k = 0; k <= n; k += 2) {
if (j + k * (1 << i) > m) break;
dp[i][j + k * (1 << i)] = (dp[i][j + k * (1 << i)] + dp[i - 1][j] * C(n, k)) % mod;
}
dl(dp[20][m]);
return 0;
}
/**
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* ┃ ┃
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* ┃ ┃ + + + +Code is far away from
* ┃ ┃ + bug with the animal protecting
* ┃ ┗━━━┓ 神兽保佑,代码无bug
* ┃ ┣┓
* ┃ ┏┛
* ┗┓┓┏━┳┓┏┛ + + + +
* ┃┫┫ ┃┫┫
* ┗┻┛ ┗┻┛+ + + +
*/