hdu 1208 pascal travel

好简单的一个题目，放在搜索专题里，大家都以为难，老师跑过来说这道题很简单，叫我做一下。。

有点变态的就是直接一个一个数据读进去没运行结果出，改成字符串读进去就可以。有点搞不懂。

Pascal's Travels

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 6   Accepted Submission(s) : 4

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Problem Description

An n x n game board is populated with integers, one nonnegative integer per square. The goal is to travel along any legitimate path from the upper left corner to the lower right corner of the board. The integer in any one square dictates how large a step away from that location must be. If the step size would advance travel off the game board, then a step in that particular direction is forbidden. All steps must be either to the right or toward the bottom. Note that a 0 is a dead end which prevents any further progress.


Consider the 4 x 4 board shown in Figure 1, where the solid circle identifies the start position and the dashed circle identifies the target. Figure 2 shows the three paths from the start to the target, with the irrelevant numbers in each removed.


Figure 1

Figure 2

Input

The input contains data for one to thirty boards, followed by a final line containing only the integer -1. The data for a board starts with a line containing a single positive integer n, 4 <= n <= 34, which is the number of rows in this board. This is followed by n rows of data. Each row contains n single digits, 0-9, with no spaces between them.

Output

The output consists of one line for each board, containing a single integer, which is the number of paths from the upper left corner to the lower right corner. There will be fewer than 2^63 paths for any board.

Sample Input

423311213123131104333212131232212051110101111111111110111101-1

Sample Output

307

#include<iostream>
using namespace std;
#define max 50
int a[max][max];
__int64 sum[max][max];
int main()
{
    int n,i,j;
    char ch[50];
    while(scanf("%d",&n)!=EOF)
    {
       getchar();
       if (n==-1) return 0;
       for(i=0;i<n;i++)
       {
         gets(ch);
         for(j=0;j<n;j++)
         a[i][j]=ch[j]-'0';
       }
       memset(sum,0,sizeof(sum));
       sum[0][0]=1;
       for (i=0;i<n;i++)
       {
         for (j=0;j<n;j++)
        {
            if (a[i][j]==0) continue;
            sum[i+a[i][j]][j]+=sum[i][j];
            sum[i][j+a[i][j]]+=sum[i][j];
        }
       }
       cout<<sum[n-1][n-1]<<endl;
    }
    return 0;
}