URAL - 1920 Titan Ruins: the Infinite Power of Magic(乱搞)
2013-08-15 20:47
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搞死人的题目,,,
就是在n*n的方格中找路径长度为L的回路。
开始的思路值适合n为偶数的情况,而忽视了奇数的case,所以wa了一次。
然后找奇数case的策略,代码从70多行变成了100多,然后改了又改,自己在下面测了好久,交上去1y,但心里却无成就感。
这样的一个题目,提不上什么思路,可以算作是乱搞的,下次比赛中再次碰到类似甚至同样的题目,我并不能保证能写出来。
代码如下:
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#include <algorithm>
#define M 1005
using namespace std;
int n, l, maxx;
int main ()
{
scanf("%d%d",&n, &l);
if(l&1||n*n<l) printf("Unsuitable device\n");
else
{
printf("Overwhelming power of magic\n");
int c = l/(2*n);
int last = l%(2*n);
printf("1 1\n");
if(l<2*n)
{
int m = last/2;
for(int i = 2; i <= m; ++i)
printf("%d %d\n",2*c+1,i);
for(int i = m; i >= 2; --i)
printf("%d %d\n",2*c+2,i);
printf("2 1\n");
}
else if(n&1)
{
maxx = 2*c;
if(last==0)
{
for(int i = 1; i <= 2*c; ++i)
{
if(i&1)
for(int j = 2; j <= n; ++j)
printf("%d %d\n",i,j);
else
for(int j = n; j >= 2; --j)
printf("%d %d\n",i,j);
}
}
else
{
for(int i = 1; i < 2*c; ++i)
{
if(i&1)
for(int j = 2; j <= n; ++j)
printf("%d %d\n",i,j);
else
for(int j = n; j >= 2; --j)
printf("%d %d\n",i,j);
}
int nn = n-2*c;
int cc = last/(2*nn);
int ll = last%(2*nn);
int mm = n-2*cc;
for(int i = n; i > n-2*cc; --i)
{
if(i&1)
for(int j = 2*c; j <= n; ++j)
printf("%d %d\n",j,i);
else
for(int j = n; j >= 2*c; --j)
printf("%d %d\n",j,i);
}
if(ll==0)
{
for(int i = mm; i >= 2; --i)
printf("%d %d\n",2*c,i);
}
else
{
int tm = ll/2;
for(int i = 2*c; i <= 2*c+tm; ++i)
printf("%d %d\n",i,mm);
for(int i = 2*c+tm; i >= 2*c; --i)
printf("%d %d\n",i,mm-1);
mm-=2;
for(int i = mm; i >= 2; --i)
printf("%d %d\n",2*c,i);
}
}
for(int i = maxx; i >= 2; --i)
printf("%d 1\n",i);
}
else
{
for(int i = 1; i <= 2*c; ++i)
{
if(i&1)
for(int j = 2; j <= n; ++j)
printf("%d %d\n",i,j);
else
for(int j = n; j >= 2; --j)
printf("%d %d\n",i,j);
}
if(last == 0)
maxx = 2*c;
else if(last == 2)
{
printf("%d %d\n",2*c+1,2);
maxx = 2*c+1;
}
else
{
int m = last/2;
maxx = 2*c+2;
for(int i = 2; i <= m; ++i)
printf("%d %d\n",2*c+1,i);
for(int i = m; i >= 2; --i)
printf("%d %d\n",2*c+2,i);
}
for(int i = maxx; i >= 2; --i)
printf("%d 1\n",i);
}
}
return 0;
}
就是在n*n的方格中找路径长度为L的回路。
开始的思路值适合n为偶数的情况,而忽视了奇数的case,所以wa了一次。
然后找奇数case的策略,代码从70多行变成了100多,然后改了又改,自己在下面测了好久,交上去1y,但心里却无成就感。
这样的一个题目,提不上什么思路,可以算作是乱搞的,下次比赛中再次碰到类似甚至同样的题目,我并不能保证能写出来。
代码如下:
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#include <algorithm>
#define M 1005
using namespace std;
int n, l, maxx;
int main ()
{
scanf("%d%d",&n, &l);
if(l&1||n*n<l) printf("Unsuitable device\n");
else
{
printf("Overwhelming power of magic\n");
int c = l/(2*n);
int last = l%(2*n);
printf("1 1\n");
if(l<2*n)
{
int m = last/2;
for(int i = 2; i <= m; ++i)
printf("%d %d\n",2*c+1,i);
for(int i = m; i >= 2; --i)
printf("%d %d\n",2*c+2,i);
printf("2 1\n");
}
else if(n&1)
{
maxx = 2*c;
if(last==0)
{
for(int i = 1; i <= 2*c; ++i)
{
if(i&1)
for(int j = 2; j <= n; ++j)
printf("%d %d\n",i,j);
else
for(int j = n; j >= 2; --j)
printf("%d %d\n",i,j);
}
}
else
{
for(int i = 1; i < 2*c; ++i)
{
if(i&1)
for(int j = 2; j <= n; ++j)
printf("%d %d\n",i,j);
else
for(int j = n; j >= 2; --j)
printf("%d %d\n",i,j);
}
int nn = n-2*c;
int cc = last/(2*nn);
int ll = last%(2*nn);
int mm = n-2*cc;
for(int i = n; i > n-2*cc; --i)
{
if(i&1)
for(int j = 2*c; j <= n; ++j)
printf("%d %d\n",j,i);
else
for(int j = n; j >= 2*c; --j)
printf("%d %d\n",j,i);
}
if(ll==0)
{
for(int i = mm; i >= 2; --i)
printf("%d %d\n",2*c,i);
}
else
{
int tm = ll/2;
for(int i = 2*c; i <= 2*c+tm; ++i)
printf("%d %d\n",i,mm);
for(int i = 2*c+tm; i >= 2*c; --i)
printf("%d %d\n",i,mm-1);
mm-=2;
for(int i = mm; i >= 2; --i)
printf("%d %d\n",2*c,i);
}
}
for(int i = maxx; i >= 2; --i)
printf("%d 1\n",i);
}
else
{
for(int i = 1; i <= 2*c; ++i)
{
if(i&1)
for(int j = 2; j <= n; ++j)
printf("%d %d\n",i,j);
else
for(int j = n; j >= 2; --j)
printf("%d %d\n",i,j);
}
if(last == 0)
maxx = 2*c;
else if(last == 2)
{
printf("%d %d\n",2*c+1,2);
maxx = 2*c+1;
}
else
{
int m = last/2;
maxx = 2*c+2;
for(int i = 2; i <= m; ++i)
printf("%d %d\n",2*c+1,i);
for(int i = m; i >= 2; --i)
printf("%d %d\n",2*c+2,i);
}
for(int i = maxx; i >= 2; --i)
printf("%d 1\n",i);
}
}
return 0;
}
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