牛客网暑期多校训练营第六场

链接：https://www.nowcoder.com/acm/contest/144/C
来源：牛客网

题目描述

Oak is given N empty and non-repeatable sets which are numbered from 1 to N.

Now Oak is going to do N operations. In the i-th operation, he will insert an integer x between 1 and M to every set indexed between i and N.

Oak wonders how many different results he can make after the N operations. Two results are different if and only if there exists a set in one result different from the set with the same index in another result.

Please help Oak calculate the answer. As the answer can be extremely large, output it modulo 998244353.

输入描述:

The input starts with one line containing exactly one integer T which is the number of test cases. (1 ≤ T ≤ 20)

Each test case contains one line with two integers N and M indicating the number of sets and the range of integers. (1 ≤ N ≤ 10

¹⁸

, 1 ≤ M ≤ 10

¹⁸

, 

)

输出描述:

For each test case, output "Case #x: y" in one line (without quotes), where x is the test case number (starting from 1) and y is the number of different results modulo 998244353.

示例1

输入

复制

2
2 2
3 4

输出

复制

Case #1: 4
Case #2: 52

题意 ： 给你N个空的集合，并且是满足集合的性质的，每次操作是将 i - N 变成一个随机的数，求最终的集合中会有多少种情况？
思路分析 ： 我们考虑这个问题，由于n, m 都非常大，因此我们考虑这个问题的时候可以这样去想，考虑一系列的操作后会有多少种颜色，假设会有K种颜色，那么方案数为 C(m, k)，将这几种颜色插入到集合中即可，由于集合具有互异性，因此集合内的元素只取决于某个数字首次出现的何处，注意 ， 第一个地方一定要有元素，因此是 K ， 剩下的 k-1 个元素插在 n-1 个余下的位置上，再全排列即可
代码示例 ：

using namespace std;
#define ll long long
const ll maxn = 1e6+5;
const ll mod = 998244353;

ll n, m;

ll qw(ll x, ll cnt){
    ll res = 1;
    
    while(cnt){
        if(cnt&1) res *= x;
        res %= mod;
        x *= x;
        x %= mod;
        cnt >>= 1;
    }
    return res;
}
ll inv[maxn];

void init() {
    for(ll i = 1; i <= 1000000; i++){
        inv[i] = qw(i, mod-2);
    }
}
int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    ll t;
    ll kas = 1;
    init();
    
    cin >> t;
    while(t--){
        cin >> n >> m;
        ll num = min(n, m); 
        ll ans = m%mod;
        ll sum = m%mod;
        n %= mod, m %= mod;
        
        for(ll i = 2; i <= num; i++){
            sum = sum*(m-i+1+mod)%mod*(n-i+1+mod)%mod*inv[i-1]%mod;
            ans = (ans+sum)%mod;
        }
        printf("Case #%lld: %lld
",kas++, ans);    
    }
    return 0;
}