uva10069-Distinct Subsequences

题目链接请戳这里

解题思路

设\(dp[i][j]\)为\(s1[1..i]\)作为\(s2[1..j]\)子串的个数。
这样我们可以写出状态转移方程$$dp[i][j]=\begin{cases} dp[i-1][j-1]+dp[i][j-1] & \text{\(s1[i]=s2[j]\)} \
dp[i][j-1] & \text{\(s1[i]!=s2[j]\)}\end{cases}$$
其实这个转移方程可以从最长公共子序列的\(dp\)数组归纳出。
还有。。。这道题要高精度。。。

代码

#include<stdio.h>
#include<string.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#define L 10110
#define N 100
#define M 110
using namespace std;

char s1[L], s2[M];
int t;

class bign {
public:
  int len, s[N];
  bign() { memset(s, 0, sizeof(s)); len=1; }
  bign(int);
  bign operator = (const char *num) {
    len = strlen(num);
    for (int i=0; i<len; i++) s[i]=num[len-i-1]-'0';
    return *this;
  }
  bign operator = (int num) {
    char s[N];
    sprintf(s, "%d", num);
    *this = s;
    return *this;
  }
  string str() const {
    string res = "";
    for (int i=0; i<len; i++) res=(char)(s[i]+'0')+res;
    if (res == "") res = "0";
    return res;
  }
  friend ostream& operator << (ostream &out, const bign &x)
  {
    out << x.str();
    return out;
  }
  bign operator + (const bign& b) const {
    bign c;
    c.len = 0;
    for (int i=0, g=0; g || i<max(len, b.len); i++) {
      int x = g;
      if (i < len) x += s[i];
      if (i < b.len) x += b.s[i];
      c.s[c.len++] = x % 10;
      g = x / 10;
    }
    return c;
  }
};

bign dp[M][L];

int main()
{
  int n1, n2;
  scanf("%d", &t);
  while (t--) {
    //    for (int i=0; i<=M; i++)
    //      for (int j=0; j<=L; j++) dp[i][j] = 0;
    scanf("%s%s", s1+1, s2+1);
    n1 = strlen(s1+1);
    n2 = strlen(s2+1);
    for (int i=0; i<=n1; i++) dp[0][i] = 1;
    for (int i = 1; i <= n2; i++)
      for (int j = 1; j <= n1; j++) {
	if (s2[i] == s1[j]) { dp[i][j] = dp[i-1][j-1]+dp[i][j-1]; }
	else {
	  dp[i][j] = dp[i][j-1];
	}
      }
    cout << dp[n2][n1] << endl;
  }
  return 0;
}