找规律
如图所示
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<iostream>
using namespace std;
int n;
int main()
{
cin>>n;
if (n%4>1)
{
cout<<"-1n";
return 0;
}
for (int i=1;i<=n/2;i++)
{
if (i%2) cout<<i+1<<' ';
else cout<<(n-i+2)<<' ';
}
int m=n/2;
if (n%4==1)
{
cout<<m+1;
if (m+1<n) cout<<' ';
m++;
}
for (int i=m+1;i<n;i++)
{
if ((i-m)%2) cout<<n-i<<' ';
else cout<<i-1<<' ';
}
if (m+1<n) cout<<(n-1);
cout<<endl;
return 0;
}
2 seconds
256 megabytes
standard input
standard output
Vova, the Ultimate Thule new shaman, wants to build a pipeline. As there are exactly n houses in Ultimate Thule, Vova wants the city to have exactly n pipes,
each such pipe should be connected to the water supply. A pipe can be connected to the water supply if there's water flowing out of it. Initially Vova has only one pipe with flowing water. Besides, Vova has several splitters.
A splitter is a construction that consists of one input (it can be connected to a water pipe) and x output pipes. When a splitter is connected to a water
pipe, water flows from each output pipe. You can assume that the output pipes are ordinary pipes. For example, you can connect water supply to such pipe if there's water flowing out from it. At most one splitter can be connected to any water pipe.
Vova has one splitter of each kind: with 2, 3, 4,
..., k outputs. Help Vova use the minimum number of splitters to build the required pipeline or otherwise state that it's impossible.
Vova needs the pipeline to have exactly n pipes with flowing out water. Note that some of those pipes can be the output pipes of the splitters.
The first line contains two space-separated integers n and k (1 ≤ n ≤ 1018, 2 ≤ k ≤ 109).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams
or the %I64dspecifier.
Print a single integer — the minimum number of splitters needed to build the pipeline. If it is impossible to build a pipeline with the given splitters, print -1.
4 3
2
5 5
1
8 4
-1
这题显然优先找大的
直到找不下去了，判断无解或者拿个与需要的分水器相等的（显然有）
因为一个k的分水器只能增加k-1条通道
所以列方程
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<iostream>
using namespace std;
#define MAXN (1000000000000000000)
#define MAXK (1000000000)
long long n,k;
long long bin_s(long long l,long long r)
{
if (n==1) return 0;
while (r-l>1)
{
long long m=(l+r)/2;
if ((m+k-1)*(k-m)>2*(n-1)) l=m;
else r=m;
}
if ((l+k-1)*(k-l)>2*(n-1)) l=r;
if ((l+k-1)*(k-l)<2*(n-1)) l--;
if (l==0) return -1;
return k-l;
}
int main()
{
cin>>n>>k;
cout<<bin_s(1,k-1)<<endl;
return 0;
}
Problem 3 银河之星(galaxy.cpp/c/pas)
数据组数不超过10.
这题就是记忆化搜索
9点染色减少状态，map记忆化
b[i][j][k]表示棋子可否从k方向到(i,j)
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<map>
using namespace std;
#define MAXN (100+10)
#define MAXK (10+10)
#define MAXDIV (4)
int k,n,m,x2,y2,a[MAXDIV][MAXDIV],b[MAXDIV][MAXDIV][9];
int c[8][2]={{0,1},{0,2},{1,0},{2,0},{1,1},{2,2},{1,2},{2,1}}; //↑↓→←↗↙↘↖
map<long long,int> h;
bool is_ok()
{
int ans=0;
for (int i=0;i<3;i++)
for (int j=0;j<3;j++)
ans=(11*ans+a[i][j]);
if (h.find(ans)!=h.end()) return 0;
return h[ans]=1;
}
bool dfs(int x)
{
/*
for (int j=3;j;j--)
{
for (int i=1;i<=3;i++)
cout<<a[i%3][j%3];
cout<<endl;
}
cout<<endl;
*/
if (!is_ok()) return 0;
if (x==1)
{
if (a[x2][y2]) return 1;
else return 0;
}
for (int i=0;i<3;i++)
for (int j=0;j<3;j++)
if (a[i][j])
{
for (int l=0;l<8;l++)
if (a[(i+c[l][0])%3][(j+c[l][1])%3]&&a[(i+2*c[l][0])%3][(j+2*c[l][1])%3]<b[(i+2*c[l][0])%3][(j+2*c[l][1])%3][l])
{
a[i][j]--;a[(i+c[l][0])%3][(j+c[l][1])%3]--;a[(i+2*c[l][0])%3][(j+2*c[l][1])%3]++;
if (dfs(x-1)) return 1;
a[i][j]++;a[(i+c[l][0])%3][(j+c[l][1])%3]++;a[(i+2*c[l][0])%3][(j+2*c[l][1])%3]--;
}
}
return 0;
}
int main()
{
freopen("galaxy.in","r",stdin);
freopen("galaxy.out","w",stdout);
while (scanf("%d%d%d%d%d",&k,&n,&m,&x2,&y2)!=EOF)
{
x2%=3;y2%=3;
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
for (int i=0;i<3;i++)
for (int j=0;j<3;j++)
{
int h=((n/3)+(i>0)*(i<=n%3)),r=((m/3)+(j>0)*(j<=m%3));
//b[i][j][8]=((n/3)+(i>0)*(i<=n%3))*((m/3)+(j>0)*(j<=m%3));
b[i][j][8]=h*r;
if (j!=0) b[i][j][0]=max(r-1,0)*h;else b[i][j][0]=r*h;
if ((m+1)%3!=j) b[i][j][1]=max(r-1,0)*h;else b[i][j][1]=r*h;
if (i!=0) b[i][j][2]=max(h-1,0)*r;else b[i][j][2]=r*h;
if ((n+1)%3!=i) b[i][j][3]=max(h-1,0)*r;else b[i][j][3]=r*h;
b[i][j][4]=max(r-(j!=0),0)*max(h-(i!=0),0);
b[i][j][5]=max(r-((m+1)%3!=j),0)*max(h-((n+1)%3!=i),0);
b[i][j][6]=max(r-((m+1)%3!=j),0)*max(h-(i!=0),0);
b[i][j][7]=max(r-(j!=0),0)*max(h-((n+1)%3!=i),0);
}
for (int i=1;i<=k;i++)
{
int x,y;
scanf("%d%d",&x,&y);
a[x%3][y%3]++;
}
/*
for (int j=3;j;j--)
{
for (int i=1;i<=3;i++)
cout<<a[i%3][j%3];
cout<<endl;
}
cout<<endl;
*/
if (dfs(k)) cout<<"Yesn";
else cout<<"Non";
h.clear();
}
return 0;
}