HDU 6397 Character Encoding (组合数学 + 容斥)

时间：2023-03-08 23:48:45
HDU 6397 Character Encoding (组合数学 + 容斥)

题意：HDU 6397 Character Encoding (组合数学 + 容斥)

析：首先很容易可以看出来使用FFT是能够做的，但是时间上一定会TLE的，可以使用公式化简，最后能够化简到最简单的模式。

其实考虑使用组合数学，如果这个 xi 没有限制，那么就是求 x1 + x2 + x3 +... xm = k，有多少非零解，隔板法很容易得到答案 C(k+m-1, m-1)，但是有限制怎么办，使用容斥，考虑有一个变量超过 n-1，两个变量超过 n-1，等等，根据集合论，很容易知道偶加，奇减。。。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#include <list>

#include <assert.h>

#include <bitset>

#include <numeric>

#define debug() puts("++++")

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a, b, sizeof a)

#define sz size()

#define be begin()

#define ed end()

#define pu push_up

#define pd push_down

#define cl clear()

#define lowbit(x) -x&x

// #define all 1,n,1

#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)

#define freopenr freopen("in.in", "r", stdin)

#define freopenw freopen("out.out", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e17;

const double inf = 1e20;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 200000 + 10;

const int maxm = 1e6 + 10;

const LL mod = 998244353LL;

const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};

const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c) {

  return r >= 0 && r < n && c >= 0 && c < m;

}

inline int readInt(){ int x;  scanf("%d", &x);  return x; }

LL fact[maxn], inv[maxn];

LL fast_pow(LL a, int n){

  LL res = 1;

  while(n){

    if(n&1)  res = res * a % mod;

    n >>= 1;

    a = a * a % mod;

  }

  return res;

}

void init(){

  fact[0] = fact[1] = 1;

  for(int i = 2; i < maxn; ++i)  fact[i] = fact[i-1] * i % mod;

  inv[maxn-1] = fast_pow(fact[maxn-1], mod - 2);

  for(int i = maxn-2; i >= 0; --i)  inv[i] = inv[i+1] * (i+1) % mod;

}

inline LL C(int n, int m){

  if(n < m)  return 0LL;

  return fact[n] * inv[m] % mod * inv[n-m] % mod;

}

inline LL G(int x, int k){

  return C(m, x) * C(k - x * n + m - 1, m - 1) % mod;

}

int main(){

  init();

  int T, k;  cin >> T;

  while(T--){

    scanf("%d %d %d", &n, &m, &k);

    if(n > k){ printf("%I64d\n", C(m+k-1, m-1));  continue; }

    LL ans = 0;

    for(int i = 0; i <= (k + m - 1) / n; ++i)

      ans = (ans + (i&1? -G(i, k) : G(i, k))) % mod;

    printf("%I64d\n", (ans%mod+mod)%mod);

  }

  return 0;

}

　　

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