[haoi2012]容易题

容易题的确很容易,公式一推就出来了。

详见代码:

#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<string>
#include<map>
#include<cstdlib>
#include<cmath>
#include<cstdio>
#include<ctime>
#include<queue>
using namespace std;
#define LL long long
#define FILE "dealing"
#define eps 1e-10
#define db double
#define pii pair<LL,LL>
#define up(i,j,n) for(int i=j;i<=n;i++)
int read(){
	int x=0,f=1,ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9')x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	return x*f;
}
const LL maxn=200000,mod=1000000007,inf=1000000007;
bool cmin(int& a,int b){return a>b?a=b,true:false;}
bool cmax(int& a,int b){return a<b?a=b,true:false;}
LL n,m,k,top=0;
pii t[maxn],d[maxn];
LL fast(LL a,LL b){
	LL ans=1;
	for(;b;b>>=1,a=(a*a)%mod)if(b&1)ans=(ans*a)%mod;
	return ans;
}
int main(){
	freopen(FILE".in","r",stdin);
	freopen(FILE".out","w",stdout);
	n=read(),m=read(),k=read();
	LL p=(n+1)*n/2%mod;
	if(!k){
		cout<<fast(p,m)<<endl;
		return 0;
	}
	up(i,1,k)t[i].first=read(),t[i].second=read();
	sort(t+1,t+k+1);
	top=0;
	up(i,1,k)if(d[top]!=t[i])d[++top]=t[i];
	k=top;top=1;
	t[1]=d[1];
	up(i,2,k){
		if(t[top].first==d[i].first)t[top].second=(t[top].second+d[i].second)%mod;
		else t[++top]=d[i];
	}
	k=top;
	LL ans=1;
	if(t[1].first!=1){ans=(ans*fast(p,t[1].first-1))%mod;}
	t[k+1].first=m+1;
	up(i,1,k){
		ans=(ans*((p-t[i].second+mod)%mod))%mod;
		ans=(ans*fast(p,t[i+1].first-t[i].first-1)%mod);
	}
	cout<<ans<<endl;
	return 0;
}

  

原文地址:https://www.cnblogs.com/chadinblog/p/6456081.html