google codejam 2009 Qualification Round Problem C code_2009 google code jam round 1c 题目-优快云博客

本文介绍了一个使用 C++ 实现的字符串匹配算法，该算法能够高效地计算一个给定字符串中与目标模式相匹配的所有子串的数量。通过递归方式更新计数数组，实现了对任意长度输入的有效处理。

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#include <iostream>
#include <vector>
#include <string>

using namespace std;

const string b("welcome to code jam");
//const string b("a");
const int MAXLEN = 501;
const int m = b.length();
int n;
string a;
int cnt[MAXLEN][20];

void InitArray()
{
for (int i=0; i<n; ++i)
  for (int j=0; j<m; ++j)
   cnt[i][j] = -1;
for (int i=0; i<=n; ++i)
  cnt[i][m] = 1;
for (int i=0; i<m; ++i)
  cnt[n][i] = 0;
}

int Count( int i, int j )
{
if ( cnt[i][j] > -1 ) return cnt[i][j];
//if ( i == n ) return cnt[i][j] = 0;

if ( a[i] == b[j] ) { return cnt[i][j] = (Count( i + 1, j + 1 ) + Count( i + 1, j )) % 10000; }
else return cnt[i][j] = Count(i+1, j);
}

int main()
{
#ifndef ONLINE_JUDGE
freopen("data.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
int N;

cin >> N;
cin.get();
for (int i=1; i<=N; ++i)
{
  char astr[MAXLEN];
  cin.getline(astr, MAXLEN);
  a = astr;
        n = a.length();
  InitArray();
  cout << "Case #"<<i << ": " ;
  int number = Count(0, 0);
  if (number < 1000) cout << 0;
  if (number < 100) cout << 0;
  if (number < 10) cout << 0;
  cout << number;
  cout << "/n";
}
return 0;
}