A knight jumps around an infinite chessboard. The chessboard is an unexplored territory. In the spirit of explorers, whoever stands on a square for the first time claims the ownership of this square. The knight initially owns the square he stands, and jumps NN times before he gets bored.
Recall that a knight can jump in 8 directions. Each direction consists of two squares forward and then one squaure sidways.
After NN jumps, how many squares can possibly be claimed as territory of the knight? As NN can be really large, this becomes a nightmare to the knight who is not very good at math. Can you help to answer this question?
Input
The first line of the input gives the number of test cases, TT. TT test cases follow.
Each test case contains only one number NN, indicating how many times the knight jumps.
1≤T≤1051≤T≤105
0≤N≤1090≤N≤109
Output
For each test case, output one line containing “Case #x: y”, where xx is the test case number (starting from 1) and yy is the number of squares that can possibly be claimed by the knight.
Sample Input
3 0 1 7
Sample Output
Case #1: 1 Case #2: 9 Case #3: 649
他们数论的做法我看不懂 但打表后发现 6以后每一层的递增量是一个等差数列 公差为28
还卡了精度 要用long long unsigned
#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
/*
int map[1000][1000];
int visit[1000][1000];
int dir[8][2]={-2,-1,-1,-2,1,-2,2,-1,-2,1,-1,2,1,2,2,1};
int ans[1000];
struct node
{
int step,x,y;
};
int main()
{
queue<node> q;
node start,in,ne;
start.x=500,start.y=500,start.step=0;
visit[500][500]=1;
ans[0]=1;
q.push(start);
while(!q.empty())
{
in=q.front();
q.pop();
if(in.step>20) break;
for(int i=0;i<8;i++)
{
ne.y=in.y+dir[i][1];
ne.x=in.x+dir[i][0];
ne.step=in.step+1;
if(!visit[ne.x][ne.y])
{
q.push(ne);
ans[ne.step]++;
visit[ne.x][ne.y]=1;
}
}
}
int sum=1,presum;
for(int i=1;i<=20;i++)
{
sum+=ans[i];
cout<<"cishu"<<ans[i]<<'\t'<<"sum"<<sum<<"cha"<<ans[i]-ans[i-1]<<endl;
presum=sum;
}
}*/
int main()
{
int ans[8]={1,9,41,109,205,325,473,649};
int t;
cin>>t;
for(int i=1;i<=t;i++)
{
unsigned long long num;
scanf("%llu",&num);
if(num<=6)
{
printf("Case #%d: %d\n",i,ans[num]);
}
else
{
unsigned long long base=473,s=num-6,ss;
base+=148*s;
ss=((1+s)*s)/2;
base+=28*ss;
printf("Case #%d: %llu\n",i,base);
}
}
}