http://www.elijahqi.win/archives/1225
— This is not playing but duty as allies of justice, Nii-chan!
— Not allies but justice itself, Onii-chan!
With hands joined, go everywhere at a speed faster than our thoughts! This time, the Fire Sisters — Karen and Tsukihi — is heading for somewhere they’ve never reached — water-surrounded islands!
There are three clusters of islands, conveniently coloured red, blue and purple. The clusters consist of a, b and c distinct islands respectively.
Bridges have been built between some (possibly all or none) of the islands. A bridge bidirectionally connects two different islands and has length 1. For any two islands of the same colour, either they shouldn’t be reached from each other through bridges, or the shortest distance between them is at least 3, apparently in order to prevent oddities from spreading quickly inside a cluster.
The Fire Sisters are ready for the unknown, but they’d also like to test your courage. And you’re here to figure out the number of different ways to build all bridges under the constraints, and give the answer modulo 998 244 353. Two ways are considered different if a pair of islands exist, such that there’s a bridge between them in one of them, but not in the other.
Input
The first and only line of input contains three space-separated integers a, b and c (1 ≤ a, b, c ≤ 5 000) — the number of islands in the red, blue and purple clusters, respectively.
Output
Output one line containing an integer — the number of different ways to build bridges, modulo 998 244 353.
Examples
Input
1 1 1
Output
8
Input
1 2 2
Output
63
Input
1 3 5
Output
3264
Input
6 2 9
Output
813023575
Note
In the first example, there are 3 bridges that can possibly be built, and no setup of bridges violates the restrictions. Thus the answer is 23 = 8.
In the second example, the upper two structures in the figure below are instances of valid ones, while the lower two are invalid due to the blue and purple clusters, respectively.
dp:好理解
我们通过题目可以知道每两种颜色之间的边其实是互不影响的
所以只考虑两种颜色的情况设dp[i][j]分别表示蓝色点有i个红色点有j个的方案数‘
那么我们知道新增一个红点的构造方法是什么
我们可以是这个点和谁都不连接 或者是新的第i个点可以和对面的j个点分别相连
#include<cstdio>
#define mod 998244353
int a,b,c,dp[5500][5500];
int main(){
// freopen("cf.in","r",stdin);
scanf("%d%d%d",&a,&b,&c);
for (int i=0;i<=5000;++i) dp[i][0]=dp[0][i]=1;
for (int i=1;i<=5000;++i)
for (int j=1;j<=5000;++j)
dp[i][j]=(dp[i-1][j]+(long long)dp[i-1][j-1]*j)%mod;
printf("%d",(long long)dp[a][b]*dp[b][c]%mod*dp[c][a]%mod);
return 0;
}组合数学：
我们还是给两列点 一列红色一列蓝色
然后那么我们第一个点的链接方案是蓝色点的个数种
第二个点的链接方案数是蓝色点个数-1种所以一共是n!种链接方案
考虑到我相当于是从n个点中选出一些点 另外一个也是 所以说最后答案还需要*组合数的
每次从红色点中选出a个点 蓝色点中选出b个点那么每次选择一共是k!种选择 然后分别乘以选出他们的组合数即可