Combination uoj#308. "Unr #2" Uoj Rescue plan __ Math related

the

Given a non-direction graph of N-point M-edges, all vertices of the graph are dyed so that the color of the ends is different for any one edge. There are k colors available.
The value of the 6 of the scheme of the stain is obtained. Solving

It's a simple question, but I didn't think of it when I did it ... I'm a good cook ...
is a classic problem that looks like NPC, and then only needs to output the answer%6 to make the problem specialized.
What's so special about it.
We consider the answer this way:
Ans=∑i=1ka (k,i) ∗[just use the I color scheme number] ans=\sum_{i=1}^k a (k,i) *[just use the number of colors I have.
We know the permutation number A (k,i) = (k−i+1) ∗ (k−i+2) ∗ ... ∗k A (k,i) = (k-i+1) * (k-i+2) *...*k
I believe you have learned in primary school: Any two consecutive integer products must have 2 this divisor, any three consecutive integer product must have 3 this divisor.
This is simple, when i≥3 i\ge 3 o'clock, A (k,i)%6=0, A (k,i) \% 6=0, do not care.
So we just need to consider 1 colors and 2 colors.
And then just do a little messing around.
Pay attention to the special situation of m=0 m=0.

#include <cstdio> #include <cstring> #include <algorithm> typedef long LL;
using namespace Std;
const int maxn=1000005,maxe=4000005;
int q,n,m,k,c[maxn],fa[maxn],res,cnt;
BOOL VIS[MAXN];
int Fir[maxn],nxt[maxe],son[maxe],tot; void Add (int x,int y) {son[++tot]=y; nxt[tot]=fir[x]; Fir[x]=tot} int GETFA (int x) {return FA[X]==X?X:FA[X]=GETFA (FA [x]);
    } void Merge (int x,int y) {X=GETFA (x); Y=getfa (y);
if (x!=y) fa[x]=y;
    } void Dfs (int x,int k) {vis[x]=true; c[x]=k;
                   for (int j=fir[x];j;j=nxt[j]) {if (!vis[son[j]]) DFS (SON[J],K^1);
    else if (c[son[j]]!= (k^1)) res=0;
    int main () {freopen ("uoj308.in", "R", stdin);
    Freopen ("Uoj308.out", "w", stdout);
    scanf ("%d", &q);       
        while (q--) {scanf ("%d%d%d", &n,&m,&k);
        for (int i=1;i<=n;i++) fa[i]=i; memset (FIR) (fir,0,sizeof);
        tot=0;
            for (int i=1;i<=m;i++) {int x,y; scanf ("%d%d", &x,&y); AddX,y); Add (y,x);
        Merge (X,y);
            } if (m==0) {Res=1; for (int i=1;i<=n-1;i++) res= (res*2)%6; res= (res-1+6)%6;
        printf ("%d\n", (k%6+ (k%6) * (K-1)%6*res%6)%6);
            else{for (int i=1;i<=n;i++) c[i]=-1; 
            memset (Vis) (vis,0,sizeof); Res=1;
            cnt=0;
            for (int i=1;i<=n;i++) if (!vis[i]) DFS (i,0), cnt+=1;
            for (int i=1;i<=cnt-1;i++) res= (res*2)%6;       
        printf ("%d\n", res* (k%6) * (K-1)%6);
} return 0;  }