结论很简单，就是把每个位置的数相加再相乘就好了。对于限制比较少，而位置比较多，所以处理出限制的位置，剩余位置用快速幂就好了。
1 #include<bits/stdc++.h> 2 #define N 100005 3 #define LL long long 4 #define inf 0x3f3f3f3f 5 #define ls c[x][0] 6 #define rs c[x][1] 7 using namespace std; 8 inline int ra() 9 { 10 int x=0,f=1; char ch=getchar(); 11 while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();} 12 while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();} 13 return x*f; 14 } 15 const LL mod=1000000007LL; 16 LL n,m,k; 17 struct node{ 18 LL x,y; 19 }a[N]; 20 bool operator == (node a, node b) 21 { 22 return a.x==b.x && a.y==b.y; 23 } 24 bool cmp(node a, node b) 25 { 26 if (a.x==b.x) return a.y<b.y; 27 return a.x<b.x; 28 } 29 LL ksm(LL x, LL p) 30 { 31 LL sum=1; 32 for (;p;p>>=1,x=x*x%mod) 33 if (p&1) sum=sum*x%mod; 34 return sum%mod; 35 } 36 int main() 37 { 38 n=ra(); m=ra(); k=ra(); 39 LL sum=(n+1)*n/2%mod; 40 for (LL i=1; i<=k; i++) 41 a[i].x=ra(),a[i].y=ra(); 42 sort(a+1,a+k+1,cmp); 43 LL ans=1; LL tot=1; 44 LL del=0; 45 for (int i=1; i<=k; i++) 46 if (a[i].x!=a[i-1].x) m--; 47 for (int i=1,last; i<=k; i=last) 48 { 49 LL now=sum; 50 for (last=i;a[last].x==a[i].x;last++) 51 if (last==i || a[last].y!=a[last-1].y) 52 now=(now-a[last].y+mod)%mod; 53 ans=ans*now%mod; 54 } 55 ans=(ans*ksm(sum,m))%mod; 56 cout<<(ans+mod)%mod; 57 return 0; 58 }