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

2014-11-23 21:58:34 · 作者: · 浏览: 5
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 
#include 
#include 
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; }