2019 ICPC Asia Nanjing Regional

链接：https://www.jisuanke.com/contest/5528/challenges

A. A Hard Problem

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=1e5+10;
int t,n;
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		int ans=(n+1)/2+1;
		printf("%d\n",ans);
	}
	return 0;
}

C. Digital Path

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=1000+10,mod=1e9+7;

int n,m;
ll a[maxn][maxn];
ll dp[maxn][maxn][5];
int dx[]={1,-1,0,0};
int dy[]={0,0,1,-1};
struct Node
{
	ll x,y,val;
	bool operator<(const Node& b) const
	{
		return val<b.val;
	}
};
void update(int x,int y,int nx,int ny)
{
	for(int i=2;i<=4;++i)
	{
		dp[nx][ny][i]+=dp[x][y][i-1];
		dp[nx][ny][i]%=mod;
	}
	dp[nx][ny][4]+=dp[x][y][4];
	dp[nx][ny][4]%=mod;
}
int main()
{
	priority_queue<Node> pq;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;++i)
		for(int j=1;j<=m;++j)
			scanf("%lld",&a[i][j]);
	for(int i=1;i<=n;++i)
		for(int j=1;j<=m;++j)
		{
			bool ok=1;
			for(int k=0;k<=3;++k)
			{
				int nx=i+dx[k];
				int ny=j+dy[k];
				if(nx>=1&&nx<=n&&ny>=1&&ny<=m&&a[i][j]+1==a[nx][ny])
					ok=0;			
			}
			pq.push({i,j,a[i][j]});
			if(ok) dp[i][j][1]=1;
		}	
	while(!pq.empty())
	{
		Node t=pq.top();
		pq.pop();
		bool ok=0;
		for(int i=0;i<=3;++i)
		{
			int nx=t.x+dx[i];
			int ny=t.y+dy[i];
			if(nx>=1&&nx<=n&&ny>=1&&ny<=m&&a[nx][ny]+1==a[t.x][t.y])
			{
				ok=1;
				update(t.x,t.y,nx,ny);
			}		
		}		
	}
	ll ans=0;
	for(int i=1;i<=n;++i)
	{
		for(int j=1;j<=m;++j)
		{
			bool f1=0,f2=0;
			for(int k=0;k<=3;++k)
			{
				int nx=i+dx[k];
				int ny=j+dy[k];
				if(nx>=1&&nx<=n&&ny>=1&&ny<=m)
				{
					if(a[i][j]==a[nx][ny]-1) f1=1;
					if(a[i][j]==a[nx][ny]+1) f2=1;
 				}
			}
			if(f1&&!f2) ans=(ans+dp[i][j][4])%mod;
		}	
	}
	printf("%lld\n",ans);
	return 0;
}

H. Prince and Princess

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=1e5+10;
int a,b,c; 
int main()
{
	scanf("%d%d%d",&a,&b,&c);
	if(a==1&&b==0&&c==0) puts("YES\n0");
	else
	{
		if(a>b+c) printf("YES\n%d\n",2*(b+c)+1);
		else puts("NO");
	}
	return 0;
}