nowcoder19989 [HAOI2012]容易题(EASY)

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题解

乘法原理

代码

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(_,__) for(_=1;_<=(__);_++)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
struct EasyMath
{
    ll prime[maxn], phi[maxn], mu[maxn];
    bool mark[maxn];
    ll fastpow(ll a, ll b, ll c)
    {
        ll t(a%c), ans(1ll);
        for(;b;b>>=1,t=t*t%c)if(b&1)ans=ans*t%c;
        return ans;
    }
    void shai(ll N)
    {
        ll i, j;
        for(i=2;i<=N;i++)mark[i]=false;
        *prime=0;
        phi[1]=mu[1]=1;
        for(i=2;i<=N;i++)
        {
            if(!mark[i])prime[++*prime]=i, mu[i]=-1, phi[i]=i-1;
            for(j=1;j<=*prime and i*prime[j]<=N;j++)
            {
                mark[i*prime[j]]=true;
                if(i%prime[j]==0)
                {
                    phi[i*prime[j]]=phi[i]*prime[j];
                    break;
                }
                mu[i*prime[j]]=-mu[i];
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
            }
        }
    }
    ll inv(ll x, ll p)  //p是素数
    {return fastpow(x%p,p-2,p);}
}em;
ll n, m, ans, k, last;
vector<pll> v;
#define mod 1000000007ll
int main()
{
    n=read(), m=read(), k=read();
    ll i, s=0;
    rep(i,k)
    {
        ll x=read(), y=read();
        v.emb( pll(x,y) );
    }
    sort(v.begin(),v.end());
    auto it=unique(v.begin(),v.end());
    v.erase(it,v.end());
    ans = 1;
    for(auto pr:v)
    {
        if(pr.first==last)s+=pr.second;
        else
        {
            if(last)
            {
                ans *= (n*(n+1)/2-s) %mod;
                ans %= mod;
                m--;
            }
            s=pr.second;
        }
        last=pr.first;
    }
    if(last)
    {
        ans *= (n*(n+1)/2-s) %mod;
        ans %= mod;
        m--;
    }
    ans = ans*em.fastpow(n*(n+1)/2,m,mod)%mod;
    printf("%lld",ans);
    return 0;
}