HDU 6397 组合数学+容斥 母函数

Character Encoding

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1473    Accepted Submission(s): 546

Problem Description

In computer science, a character is a letter, a digit, a punctuation mark or some other similar symbol. Since computers can only process numbers, number codes are used to represent characters, which is known as character encoding. A character encoding system establishes a bijection between the elements of an alphabet of a certain size n and integers from 0 to n−1. Some well known character encoding systems include American Standard Code for Information Interchange (ASCII), which has an alphabet size 128, and the extended ASCII, which has an alphabet size 256.

For example, in ASCII encoding system, the word wdy is encoded as [119, 100, 121], while jsw is encoded as [106, 115, 119]. It can be noticed that both 119+100+121=340 and 106+115+119=340, thus the sum of the encoded numbers of the two words are equal. In fact, there are in all 903 such words of length 3 in an encoding system of alphabet size 128 (in this example, ASCII). The problem is as follows: given an encoding system of alphabet size n where each character is encoded as a number between 0 and n−1 inclusive, how many different words of length m are there, such that the sum of the encoded numbers of all characters is equal to k?

Since the answer may be large, you only need to output it modulo 998244353.

 

Input

The first line of input is a single integer T (1≤T≤400), the number of test cases.

Each test case includes a line of three integers n,m,k (1≤n,m≤105,0≤k≤105), denoting the size of the alphabet of the encoding system, the length of the word, and the required sum of the encoded numbers of all characters, respectively.

It is guaranteed that the sum of n, the sum of m and the sum of k don't exceed 5×106, respectively.

 

Output

For each test case, display the answer modulo 998244353 in a single line.

 

Sample Input

4

2 3 3

2 3 4

3 3 3

128 3 340

 

Sample Output

1

0

7

903

 

容斥写法

x1+x2+...+xm = k (xi>=0) 共有C(k+m-1,m-1) 种 插板法

如果有c个违反条件 把每一个违反条件的x减去n

x1'+x2'+x3'+x4'+x5'+...+xn'= k-c*n xi>=0 共有 C(k-c*n+m-1,m-1)种
　　  容斥系数 　　 变量选法
ans  = （-1)^c 　 * 　 C(m,c)       *     C(k-cn+m-1,m-1)

母函数写法

1+x+x^2+...+x^(n-1)=(1-x^n)/(1-x)

(1+x+x^2+...+x^(n-1))^m

=(1-x^n)^m/(1-x)^m
=(1-x^n)^m*(1-x)^(-m)
=(1-x^n)^m*(sum_ (x^i)*C(m+i-1,m-1)) //上篇博客说的核武器。。。。

ans=x^k 的系数
左边二项式展开 按照每个i 右边应该有k-ni
ans= sum (-1)^i*C(m,i)*C(m+k-n*I-1,m-1)

左边 x^n*i 　　　　 右边x^(k-n*i)
系数(-1)^i*C(m,i) 　 系数C(m+k-n*I-1,m-1)

AC代码

#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define all(a) (a).begin(), (a).end()
#define fillchar(a, x) memset(a, x, sizeof(a))
#define huan printf("\n");
#define debug(a,b) cout<<a<<" "<<b<<" "<<endl;
using namespace std;
const int maxn= 3e5+;
const int inf = 0x3f3f3f3f,mod=;
typedef long long ll;
ll fac[maxn],inv[maxn];
void init()
{
    fac[]=fac[]=;
    inv[]=inv[]=;
    for(ll i=;i<maxn;i++)
    {
        fac[i]=fac[i-]*i%mod;
        inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
    }
    for(ll i=;i<maxn;i++)
        inv[i]=inv[i-]*inv[i]%mod;
}
ll C(ll x,ll y)
{
    if(y>x) return ;
    if(y==||x==) return ;
    return fac[x]*inv[y]%mod*inv[x-y]%mod;
}
int main()
{
    ll n,m,k,t;
    init();
    cin>>t;
    while(t--)
    {
        cin>>n>>m>>k;
        if(k==)
        {
            cout<<<<endl;
            continue;
        }
        else if((n-)*m<k)
        {
            cout<<<<endl;
            continue;
        }
        int c=min(k/n,m);
        ll ans=;
        for(int i=;i<=c;i++)
        {
            if(i%==)
                ans=(ans+C(m,i)*C(k-i*n+m-,m-)%mod)%mod;
            else
                ans=(ans-C(m,i)*C(k-i*n+m-,m-)%mod+mod)%mod;
        }
        cout<<ans<<endl;
    }
}