洛谷P1002 过河卒

这题的标签坑了。。。其实longlong就可以，先确定马能控制的位置，然后再一个一个格子的计算。F（i，j）=f（i-1,j）+f(i,j-1).

#include <iostream>  
using namespace std;  
long long x1,y1,x2,y2,p[22][22];  
int pan(long long x1,long long y1, long long x2, long long y2)  
{  
p[x2][y2]=-1;  
if (x2+2<=x1&&y2+1<=y1) p[x2+2][y2+1]=-1;      
if (x2+2<=x1&&y2-1>=1) p[x2+2][y2-1]=-1;      
if (x2+1<=x1&&y2+2<=y1) p[x2+1][y2+2]=-1;      
if (x2+1<=x1&&y2-2>=1) p[x2+1][y2-2]=-1;      
if (x2-1>=1&&y2+2<=y1) p[x2-1][y2+2]=-1;      
if (x2-1>=1&&y2-2>=1) p[x2-1][y2-2]=-1;      
if (x2-2>=1&&y2+1<=y1) p[x2-2][y2+1]=-1;      
if (x2-2>=1&&y2-1>=1) p[x2-2][y2-1]=-1;      
return 0;  
}  
int main()  
{  
 cin >>x1>>y1>>x2>>y2;  
x2++; y2++; x1++; y1++;  
p[1][0]=1;  
pan(x1,y1,x2,y2);  

for (long long o1=1;o1<=x1;o1++)  
for (long long o2=1;o2<=y1;o2++)  
{if (p[o1][o2]==-1)p[o1][o2]++; else p[o1][o2]=p[o1-1][o2]+p[o1][o2-1];}  
cout <<p[x1][y1];  
return 0;  
}