牛客网 Generation I

标题：C、Generation I | 时间限制：3 秒 | 内存限制：256M 
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 ≤ 1018, 1 ≤ M ≤ 1018
, )
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
2
2 2
3 4
输出
Case #1: 4
Case #2: 52

AC代码

#include <iostream>
#include<bits/stdc++.h>
using namespace std;
#define mod 998244353
#define N 1000003
long long inv[N];
int main()
{
    std::ios::sync_with_stdio(false);
    inv[0]=inv[1]=1;
    for(int i = 2; i < N; ++i)
        inv[i] = (mod - mod/i) * inv[mod%i] % mod;
    int t;
    cin >> t;
    for(int ca = 1; ca <= t; ++ca)
    {
        long long n, m;
        cin >> n >> m;
        long long minn = min(n, m);
        long long ans = 0;
        long long sum = 1;
        for(int k = 1; k <= minn; ++k)
        {
            long long temp1 = (m - k + 1) % mod;
            long long temp2 = (n - k + 1) % mod;
            if (m >= m - k +1)
                sum = (sum * temp1) % mod;
            if (n - 1 >= n - k + 1)
                sum = (sum * temp2) % mod;
            sum = (sum * inv[k-1]) % mod;
            ans = (ans + sum) % mod;
        }
        cout << "Case #" << ca << ": " << ans << endl;
    }
    return 0;
}

O(N) 求连续逆元

O(n)求逆元

inv[i] = ( MOD - MOD / i ) * inv[MOD%i] % MOD

证明：

设t = MOD / i , k = MOD % i

则有 t * i + k == 0 % MOD

有 -t * i == k % MOD

两边同时除以ik得到

-t * inv[k] == inv[i] % MOD

即

inv[i] == -MOD / i * inv[MOD%i]

即

inv[i] == ( MOD - MOD / i) * inv[MOD%i]

证毕

适用于MOD是质数的情况，能够O(n)时间求出1~n对模MOD的逆

代码实现：

    inv[1]=1;  
    for (int i=2;i<=n;++i)  
        inv[i]=(ll)(mod-mod/i)*inv[mod%i]%mod;  

O(N) 求阶乘逆元

fac[0]=fac[1]=1;

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

inv[MAXN]=quipow(fac[MAXN],mod-2);

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