P2220 [HAOI2012] easy questions

P2220 [HAOI2012] easy questions

Provided $ t = \ frac {n (n + 1)} {2} $

$ K = 0 $, the apparent $ ans = m ^ t $

Consider only one position $ x $ unavailable number $ y $, $ x $ position is the contribution to the total volume of $ ty $

In this case $ ans = (m-1) ^ t * (ty) $

So we put all the weight position to look

Quick statistical power of good location, some crippled direct enumeration position statistics

#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
int read(){
    char c=getchar(); int x=0;
    while(c<'0'||c>'9') c=getchar();
    while('0'<=c&&c<='9') x=x*10+c-48,c=getchar();
    return x;
}
const ll P=1e9+7;
int k,tk,m; ll n,t,ans;
struct Lim{int x;ll y;}a[100005];
int cmp(Lim A,Lim B){return A.x==B.x?A.y<B.y:A.x<B.x;}
ll Pow(ll x,ll y){ll re=1; for(;y;y>>=1,x=x*x%P)if(y&1)re=re*x%P; return re;}
int main(){
    n=read(); m=read(); k=read(); t=n*(n+1)/2;
    for(int i=1;i<=k;++i) a[i].x=read(),a[i].y=read();
    sort(a+1,a+k+1,cmp);
    for(ll i=1,s=0;i<=k;++i){
        if(a[i].x==a[i+1].x){
            if(a[i].y!=a[i+1].y) s+=a[i].y;
        }else a[i].y+=s,a[++tk]=a[i],s=0;
    }ans=Pow(t%P,m-tk);
    for(int i=1;i<=tk;++i) ans=(t-a[i].y)%P*ans%P;
    printf("%lld",ans);
    return 0;
}