递推水题 Codeforces Round #289 (Div. 2, ACM ICPC Rules) A. Maximum in Table

题目传送门

 1 /*
 2     模拟递推水题
 3 */
 4 #include <cstdio>
 5 #include <iostream>
 6 #include <cmath>
 7 #include <algorithm>
 8 #include <cstring>
 9 #include <string>
10 #include <map>
11 using namespace std;
12 
13 const int MAXN = 10 + 10;
14 const int INF = 0x3f3f3f3f;
15 int a[MAXN][MAXN];
16 
17 int main(void)
18 {
19     #ifndef ONLINE_JUDGE
20     freopen ("A.in", "r", stdin);
21     #endif
22 
23     int n;
24     while (~scanf ("%d", &n))
25     {
26         for (int i=1; i<=n; ++i)
27         {
28             a[1][i] = 1;
29             a[i][1] = 1;
30         }
31 
32         for (int i=2; i<=n; ++i)
33         {
34             for (int j=2; j<=n; ++j)
35             {
36                 a[i][j] = a[i-1][j] + a[i][j-1];
37             }
38         }
39 
40         printf ("%d
", a[n][n]);
41     }
42 
43     return 0;
44 }
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原文地址:https://www.cnblogs.com/Running-Time/p/4366690.html