題目連接：10069 - Distinct Subsequences

題目大意：給出兩個字符串x （lenth < 10000), z (lenth < 100), 求在x中有多少個z。

解題思路：二維數組DP， 有類似于求解最長公共子序列， cnt[i][j]表示在x的前j個字符中有多少個z 前i個字符。

狀態轉移方程 ?

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];

計算的時候使用高精度， 并且要見j == 0的情況歸1， i == 0 的情況歸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() {	// 刪除前導0.

	bign now = change(*this);

	while (now.s[now.len - 1] == 0 && now.len > 1) {

	    now.len--;

	}

	*this = change(now);

    }



    void put() {    // 輸出數值。

	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;

}


  

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uva 10069 Distinct Subsequences(高精度 + DP求解子串個數）