Description
Input
Output
Sample Input
Sample Output
解題思路:
這題徹底考察了對威佐夫博弈的掌握程度。詳見http://blog.csdn.net/userluoxuan/article/details/38336287
AC代碼:
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
const int maxn = 500005;
int a[maxn], b[maxn], pos = 0;
void Func()
{
for(int i = 0;;i++)
{
a[i] = (int)(i * ((sqrt(5.0) + 1) / 2));
b[i] = a[i] + i;
if(b[i] > 1000000)
{
pos = i;
break;
}
}
}
int Find(int num)
{
int low = 0, high = pos, mid;
while(high >= low)
{
mid = (high + low) / 2;
if(a[mid] == num)
return mid;
if(a[mid] > num)
high = mid - 1;
if(a[mid] < num)
low = mid + 1;
}
return -1;
}
int Find2(int num)
{
int low = 0, high = pos, mid;
while(high >= low)
{
mid = (high + low) / 2;
if(b[mid] == num)
return mid;
if(b[mid] > num)
high = mid - 1;
if(b[mid] < num)
low = mid + 1;
}
return -1;
}
int main()
{
int m, n;
Func();
while(scanf("%d%d", &m, &n) && m && n)
{
int k = n - m;
if((int)(k * ((sqrt(5.0) + 1) / 2)) == m)
printf("0\n");
else
{
printf("1\n");
printf("%d %d\n", a[n - m], b[n - m]);
int j = Find(m);
if(j >= 0 && n > b[j])
printf("%d %d\n", a[j], b[j]);
if(j < 0)
{
int l = Find2(m);
printf("%d %d\n", a[l], m);
}
}
}
return 0;
}