Generation I--组合数和数学

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

Generation I

时间限制：C/C++ 3秒，其他语言6秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

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 ≤ 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

复制

2
2 2
3 4

输出

复制

Case #1: 4
Case #2: 52

C呐就是个排列组合，题意很难懂，反正四个人没看懂，可能时候因为读了几遍不理解，然后看目前AC的不多，就导致了刚刚说的四个人的状态，所以先说一下题意，给出N个集合，有N个操作，第i次操作从1~M中选择一个数，填入i~N的集合中。举个栗子，N=2,M=2

第一次操作选1，填入N1~N2集合中，第二次选1填入N2~N2中

N1: 1

N2:1 1 就是 1 1就是说集合中如果有相同的数字就保留一个。

第一次操作选1，填入N1~N2集合中，第二次选2填入N2~N2中

N1: 1

N2:1 2 就是 1 1 2就是说集合中如果有相同的数字就保留一个。

第一次操作选2，填入N1~N2集合中，第二次选1填入N2~N2中

N1: 2

N2:2 1 就是2 2 1就是说集合中如果有相同的数字就保留一个。

第一次操作选2，填入N1~N2集合中，第二次选2填入N2~N2中

N1: 2

N2:2 2 就是2 2 1就是说集合中如果有相同的数字就保留一个。

总共四种结果。

怎么考虑？

首先选一个数填就是A(m,1)种情况，1 2和2 1不同

选两个数填就是A(m,2)种情况，例如一开始集合都放1 1 1 1 1 1 1...

那么2放到哪里？以1 12 12 12 12 12...为例，不论以后放1还是2最终都是这个结果。所以问题就转化成了2一开始放到了哪里？

应用隔板法，每个隔板就是一开始放2的位置。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL MOD = 998244353;
const LL MAXN = 2e6+888;
LL Fac[MAXN];
LL RFac[MAXN];
LL FastPOW(LL a, LL k) {LL ret = 1;while(k) {if (k&1) {ret = (ret * a) % MOD;}a = (a * a) % MOD;k = (k >> 1);}return ret;
}
void Init() {Fac[0] = 1;for (LL i=1; i<MAXN; ++i) {Fac[i] = Fac[i-1] * i % MOD;}RFac[MAXN-1] = FastPOW(Fac[MAXN-1], MOD-2);for (LL i=MAXN-2; ~i; --i) {RFac[i] = RFac[i+1] * (i+1) % MOD;}
}
int main()
{Init();int T;scanf("%d", &T);LL n, m;for (int cas = 1; cas <= T; ++cas){scanf("%lld %lld", &n, &m);LL tmp1=Fac[m]% MOD;LL tmp2=Fac[n-1]%MOD;LL res = 0;for (LL i=1; i<=min(n,m); ++i){res=(res+((((tmp1*tmp2)%MOD*RFac[m-i])%MOD*RFac[n-i])%MOD*RFac[i-1])%MOD)%MOD;}printf("Case #%d: %lld\n", cas, res);}return 0;
}

直接按公式写代码会发生段错误，因为你预处理不了1e18。那怎么办呢？

化简公式。展开

#include<bits/stdc++.h>using namespace std;
typedef long long ll;
const int p = 998244353;
const int maxn = 1e6+1000;
ll inv[maxn];
ll N, M;int main()
{int T;cin >> T;inv[0] = inv[1] = 1;for(int i=2;i<maxn;++i)inv[i]=(p-p/i)*inv[p%i]%p;for(int cas=1;cas <= T;++cas){cin >> N >> M;ll ans = 0, f = M%p, gg = 1;for(int i=1;i<=N&&i<=M;++i){ans=(ans+f*gg)%p;f=f*((M-i)%p)%p;gg=gg*((N-i)%p)%p*inv[i]%p;}printf("Case #%d: %lld\n", cas, ans);}return 0;
}