USACO 2.3 Cow Pedigrees

最新推荐文章于 2025-09-04 16:45:21 发布

转载最新推荐文章于 2025-09-04 16:45:21 发布 · 38 阅读

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原文链接：http://www.cnblogs.com/huanyan/archive/2011/05/06/USACO_Cow_Pedigrees.html

文章标签：

#c/c++

本文提供了一段使用 C++ 编写的程序代码，该程序解决了 USACO 中关于牛谱系的问题。通过读取输入文件并进行特定的数学运算，程序能够有效地计算出所需的答案，并将结果写入输出文件。

TASK: nocows
LANG: C++

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.000 secs, 3104 KB]
   Test 2: TEST OK [0.000 secs, 3104 KB]
   Test 3: TEST OK [0.000 secs, 3104 KB]
   Test 4: TEST OK [0.000 secs, 3104 KB]
   Test 5: TEST OK [0.000 secs, 3104 KB]
   Test 6: TEST OK [0.000 secs, 3104 KB]
   Test 7: TEST OK [0.000 secs, 3104 KB]
   Test 8: TEST OK [0.000 secs, 3104 KB]
   Test 9: TEST OK [0.000 secs, 3104 KB]
   Test 10: TEST OK [0.000 secs, 3104 KB]
   Test 11: TEST OK [0.000 secs, 3104 KB]
   Test 12: TEST OK [0.000 secs, 3104 KB]

All tests OK.

1 /*
2 PROG: nocows
3 ID: jiafeim1
4 LANG: C++
5 */
6
7 #include <algorithm>
8 #include <iostream>
9 #include <fstream>
10
11 using namespace std;
12
13 int res[104][204]={0};
14 int main()
15 {
16 std::ifstream fin("nocows.in");
17 std::ofstream fout("nocows.out");
18
19 int n,k;
20 fin>>n>>k;
21 for(int i = 1 ;i<=k;++i)
22 {
23 res[i][0] = 1;
24 res[i][1] = 1;
25 }
26
27 for(int l = 1;l<=k;++l)
28 {
29 for(int num =1;num<=n;num+=2)
30 {
31 for(int i = 1;i<=num-1;i+=2)
32 {
33 res[l][num]+=(res[l-1][i]*res[l-1][num-1-i])%9901;
34 }
35 res[l][num] %=9901;
36 }
37 }
38 int temp = res[k][n]-res[k-1][n];
39 if(temp<0)
40 temp+=9901;
41 else
42 temp%=9901;
43 fout<<temp<<endl;
44 fin.close();
45 fout.close();
46
47 }

转载于:https://www.cnblogs.com/huanyan/archive/2011/05/06/USACO_Cow_Pedigrees.html