Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3146 Accepted Submission(s): 1852
思路:从右下角开始算出每个位置到终点的方法数,计算的方法是在有限的步数里把所有能走的都走一遍,加上它们的值。题目的表达很有问题,含糊不清,这题不太好,不再赘述。
1 #include<stdio.h> 2 #include<stdlib.h> 3 #include<string.h> 4 #define MAX 105 5 #define MOD 10000 6 7 int main(void) 8 { 9 int t,n,m; 10 int s[MAX][MAX],dp[MAX][MAX]; 11 12 scanf("%d",&t); 13 while(t --) 14 { 15 memset(dp,0,sizeof(dp)); 16 17 scanf("%d%d",&n,&m); 18 for(int i = 1;i <= n;i ++) 19 for(int j = 1;j <= m;j ++) 20 scanf("%d",&s[i][j]); 21 22 dp[n][m] = 1; 23 for(int i = n;i >= 1;i --) 24 for(int j = m;j >= 1;j --) 25 { 26 if(i == n && j == m) 27 continue; 28 else 29 for(int r = 0;r <= s[i][j] && i + r <= n;r ++) 30 for(int c = 0;c + r <= s[i][j] && j + c <= m;c ++) 31 dp[i][j] = (dp[i][j] + dp[i + r][j + c]) % MOD; 32 } 33 printf("%d\n",dp[1][1]); 34 } 35 36 return 0; 37 }
原文:http://www.cnblogs.com/xz816111/p/4172902.html