Character Encoding

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 510    Accepted Submission(s): 194
 

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

題意：很明確的題意， 就是要求0到n-1中選出m個數， 使其和恰好為k的方案數

思路：首先我們先不考慮n， 先考慮m個數， 組成的和為k的組合方式， 我們可以理解成隔板法放小球問題， 將k看作k個1然後將其分成m份， 可以為空， 這樣我們就可以看作隔板法放小球了， 答案為C（k + m - 1， m - 1）； 現在我們求出來的是不管n的大小的， 顯然當k>=n的時候會出現某一份中超過n的組合， 我們可以找出是哪一份， 因為這m份每份都有可能是超出n的那一個， 所以是C（m， i）， i表示超出n的份數， 這樣我們只是找出了在份數固定的情況下哪幾個份中可能出現大於n的情況， 但是組合情況還是沒有算， 我們考慮之前的隔板放小球問題， 所以我們直接另k為k-i * n就可以了， 這樣C（m， i） * C（k - i * n + m - 1， m - 1）就表示在有i份超出n的情況下的所有組合情況了， 然後我們就可以用容斥去解決了。

程式碼：

#include <iostream>
#include <cstdio>

#define MOD 998244353
const int maxed = 100000 + 10;
typedef unsigned long long ll;

ll A[maxed * 2];
int n, m, k;

int main()
{
    ll slove(ll w, int x);
    A[0] = 1;
    for (int i = 1; i < 2 * maxed; ++i)
        A[i] = A[i - 1] * i % MOD;
    int N;
    scanf("%d", &N);
    while (N--) {
        scanf("%d%d%d", &n, &m, &k);
        ll answer = A[k + m - 1] * slove(A[m - 1] * A[k] % MOD, MOD - 2) % MOD;
        //std::cout << "=====" << answer << std::endl;
        for (int i = 1; ; ++i) {
            if (1LL * i * n > k || i > m)
                break;
            if (i % 2)
                answer = (answer - A[m] * slove(A[i] * A[m - i] % MOD, MOD - 2) % MOD * A[k - i * n + m - 1] % MOD * slove(A[m - 1] * A[k - i * n] % MOD, MOD - 2) % MOD + MOD) % MOD;
            else
                answer = (answer + A[m] * slove(A[i] * A[m - i] % MOD, MOD - 2) % MOD * A[k - i * n + m - 1] % MOD * slove(A[m - 1] * A[k - i * n] % MOD, MOD - 2) % MOD) % MOD;
        }
        printf("%lld\n", answer);
    }
    return 0;
}

ll slove(ll w, int x)
{
    ll a = 1;
    while (x) {
        if (x & 1)
            a = a * w % MOD;
        x >>= 1;
        w = w * w % MOD;
    }
    return a;
}