UVA 10069 Distinct Subsequences

​​http://uva.onlinejudge.org/external/100/10069.pdf​​

两个字符串，求下一个在上一个之中出现的次数。

如果a[i] == b[j]则 dp[i][j] = dp[i-1][j] + dp[i-1][j-1]

如果a[i] != b[j]则 dp[i][j] = dp[i-1][j]

这里还涉及到高精度加法，是时候使用C语言的类了。

#include<iostream>  
#include<algorithm>
#include<math.h>
#include<cstdio>
#include<string>
#include<string.h>
using namespace std;
const int maxn = 30;
const int base = 100000000;
struct bignum
{
  int len, s[maxn];
  bignum() { memset(s, 0, sizeof(s)); len = 1; }
  bignum operator = (const bignum &b)
  {
    len = b.len;
    memcpy(s, b.s, maxn*sizeof(int));
    return *this;
  }
};

bignum operator +(const bignum &a, const bignum &b)
{
  bignum c;
  int jw = 0, i;
  for (i = 0; i < max(b.len, a.len) || jw>0; i++)
  {
    if (i < a.len) jw += a.s[i];
    if (i < b.len) jw += b.s[i];
    c.s[i] = jw % base;   jw /= base;
  }
  c.len = i;    return c;
}

int main(){
  char s[20000], a[200];
  int t;
  cin >> t;
  while (t--)
  {
    bignum f[200];
    scanf("%s%s", s, a);
    int k = strlen(a) - 1;
    f[0].s[0] = 1;
    for (int i = 0; s[i] != '\0'; i++)
      for (int j = k; j >= 0; j--)
        if (a[j] == s[i]) f[j + 1] = f[j + 1] + f[j];

    printf("%d", f[k + 1].s[f[k + 1].len - 1]);
    for (int i = f[k + 1].len - 2; i >= 0; i--) printf("%08d", f[k + 1].s[i]);
    cout << endl;
  }
  return 0;
}