• 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
    ");
    		else
    		{
    			for (int i=1;i<k;i++) printf("%d ",ans[i]);
    			printf("%d
    ",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。

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  • 原文地址:https://www.cnblogs.com/Gloid/p/10936526.html
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