Comet OJ Contest 4

　　A：签到。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int T,a[6],cnt[6];
signed main()
{
	T=read();
	while (T--)
	{
		for (int i=1;i<=5;i++) a[i]=read();
		memset(cnt,0,sizeof(cnt));
		for (int i=1;i<=5;i++) cnt[a[i]]++;
		int mx=0;
		for (int i=1;i<=5;i++) if (cnt[i]>cnt[mx]) mx=i;
		cout<<mx<<endl;
	}
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　B：k是奇数时函数值均为1，k是偶数时每k+1个出现一个0。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
ll read()
{
	ll x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int T;
ll l,r,k;
ll f(ll x)
{
	return x-(x+1)/(k+1);
}
signed main()
{
	T=read();
	while (T--)
	{
		l=read(),r=read(),k=read();
		if (k&1) cout<<r-l+1<<endl;
		else cout<<f(r)-f(l-1)<<endl;
	}
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　C：暴力枚举横竖各切多少刀，将矩阵压成一行可以得到该情况下列的划分位置，压成一列可以得到该情况下行的划分位置，然后二维前缀和暴力验证即可。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
#define N 1010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
ll read()
{
	ll x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int T,n,m,k,tot,a[N][N],s[N],s2[N],s3[N][N],ans[N<<1],way[N<<1];
int calc(int x,int y,int l,int r)
{
	return s3[y][r]-s3[x-1][r]-s3[y][l-1]+s3[x-1][l-1];
}
signed main()
{
	T=read();
	while (T--)
	{
		n=read(),m=read(),k=read();tot=0;
		for (int i=1;i<=n;i++)
			for (int j=1;j<=m;j++)
			{
				char c=getc();
				if (c=='0') a[i][j]=0;else a[i][j]=1,tot++;
			}
		for (int i=1;i<=n;i++)
		{
			s[i]=s[i-1];
			for (int j=1;j<=m;j++)
			if (a[i][j]) s[i]++;
		}
		for (int i=1;i<=m;i++)
		{
			s2[i]=s2[i-1];
			for (int j=1;j<=n;j++)
			if (a[j][i]) s2[i]++;
		}
		for (int i=1;i<=n;i++) 
			for (int j=1;j<=m;j++)
			s3[i][j]=s3[i-1][j]+s3[i][j-1]-s3[i-1][j-1]+1-a[i][j];
		tot=n*m-tot;
		bool isac=0;
		for (int i=1;i<=k;i++) ans[i]=2010;
		for (int i=0;i<=k;i++)
		if (tot%((i+1)*(k-i+1))==0)
		{
			int sum=tot/((i+1)*(k-i+1));
			int last=0;
			int cnt=0;
			bool flag=1;
			for (int j=1;j<n;j++)
			if (cnt<i)
			{
				if ((j-last)*m-(s[j]-s[last])==sum*(k-i+1))
				{
					way[++cnt]=j;
					last=j;
				}
				else if ((j-last)*m-(s[j]-s[last])>sum*(k-i+1)) {flag=0;break;}
			}
			if (cnt<i) flag=0;
			if (flag)
			{
				last=0;
				for (int j=1;j<m;j++)
				if (cnt<k)
				{
					if ((j-last)*n-(s2[j]-s2[last])==sum*(i+1))
					{
						way[++cnt]=n+j-1;
						last=j;
					}
					else if ((j-last)*n-(s2[j]-s2[last])>sum*(i+1)) {flag=0;break;}
				}
			}
			if (cnt<k) flag=0;
			if (flag)
			{
				bool u=1;
				for (int x=1;x<=i;x++)
				{
					for (int y=i+1;y<=k;y++)
					{
						if (calc(way[x-1]+1,way[x],(y==i+1?1:way[y-1]-n+2),way[y]-n+1)!=sum) {u=0;break;}
					}
					if (!u) break;
				}
				if (u)
				{	
					isac=1;
					for (int j=1;j<=k;j++)
					{
						if (way[j]<ans[j])
						{
							for (int x=1;x<=k;x++) ans[x]=way[x];
							break;
						}
						else if (way[j]>ans[j]) break;
					}
				}
			}
		}
		if (!isac) printf("Impossible\n");
		else
		{
			for (int i=1;i<k;i++) printf("%d ",ans[i]);
			printf("%d\n",ans[k]);
		}
	}
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　D：打表可知10k~10k+9范围内0~9各出现一次，于是只需要计算零散部分。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1000000010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
ll read()
{
	ll x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
ll l,r,a[30],f[1000000];
int ff(ll n)
{
	if (n<1000000) return f[n];
	int cnt=0;
	while (n) a[++cnt]=n%10,n/=10;
	for (int j=1;j<cnt;j++) a[j]+=a[j+1];
	a[cnt]=0;
	ll t=0;
	for (int j=cnt;j>=1;j--) t=t*10+a[j]%10;
	return ff(t);
}
ll calc(ll n)
{
	int s=0;int w=ff(n/10*10);
	for (ll i=n/10*10;i<=n;i++) s+=(w+(i-n/10*10))%10;
	return n/10*45+s;
}
signed main()
{
	for (int i=0;i<=9;i++) f[i]=i;
	for (int i=10;i<=1000000;i++)
	{
		ll cnt=0,x=i;
		while (x) a[++cnt]=x%10,x/=10;
		for (int j=1;j<cnt;j++) a[j]+=a[j+1];
		a[cnt]=0;
		ll t=0;
		for (int j=cnt;j>=1;j--) t=t*10+a[j]%10;
		f[i]=f[t];
	}
	int T=read();
	while (T--)
	{
		l=read(),r=read();
		cout<<calc(r)-calc(l-1)<<endl;
	}
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　先咕着。