uva 10069 Distinct Subsequences(高精度 + DP求解子串个数）

题目连接：10069 - Distinct Subsequences

题目大意：给出两个字符串x （lenth < 10000), z (lenth < 100), 求在x中有多少个z。

解题思路：二维数组DP， 有类似于求解最长公共子序列， cnt[i][j]表示在x的前j个字符中有多少个z 前i个字符。

状态转移方程  

1、x[j] != z[i]              cnt[i][j] = cnt[i][j - 1];

2、x[j] == z[i]   cnt[i][j] = cnt[i][j - 1] + cnt[i - 1][j - 1];

计算的时候使用高精度， 并且要见j == 0的情况归1， i == 0 的情况归0。

#include <stdio.h>#include <string.h>#include <iostream>using namespace std;const int N = 10005;const int M = 105;struct bign {    int len, sex;    int s[M];    bign() {this -> len = 1;this -> sex = 0;memset(s, 0, sizeof(s));    }    bign operator = (const char *number) {int begin = 0;len = 0;sex = 1;if (number[begin] == '-') {    sex = -1;    begin++;}else if (number[begin] == '+')    begin++;for (int j = begin; number[j]; j++)    s[len++] = number[j] - '0';    }    bign operator = (int number) {char string[N];sprintf(string, "%d", number);*this = string;return *this;    }    bign (int number) {*this = number;}    bign (const char* number) {*this = number;}    bign change(bign cur) {bign now;now = cur;for (int i = 0; i < cur.len; i++)    now.s[i] = cur.s[cur.len - i - 1];return now;    }    void delZore() {// 删除前导0.bign now = change(*this);while (now.s[now.len - 1] == 0 && now.len > 1) {    now.len--;}*this = change(now);    }    void put() {    // 输出数值。delZore();if (sex < 0 && (len != 1 || s[0] != 0))    cout << "-";for (int i = 0; i < len; i++)    cout << s[i];    }    bign operator + (const bign &cur){  bign sum, a, b;  sum.len = 0;a = a.change(*this);b = b.change(cur);for (int i = 0, g = 0; g || i < a.len || i < b.len; i++){      int x = g;      if (i < a.len) x += a.s[i];      if (i < b.len) x += b.s[i];      sum.s[sum.len++] = x % 10;      g = x / 10;  }  return sum.change(sum);      } };bign cnt[M][N], sum;char x[N], z[M];int main() {    int cas;    scanf("%d", &cas);    while (cas--) {scanf("%s%s", x, z);int n = strlen(x), m = strlen(z);for (int i = 0; i <= n; i++)    cnt[0][i] = 1;for (int i = 1; i <= m; i++) {    cnt[i][0] = 0;    for (int j = 1; j <= n; j++) {cnt[i][j] = cnt[i][j - 1];if (z[i - 1] == x[j - 1])       cnt[i][j] = cnt[i][j] + cnt[i - 1][j - 1];    }}cnt[m][n].put();printf("\n");    }    return 0;}