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("
");
    }
    return 0;
}