uva10069 - Distinct Subsequences(大数+DP）

题目：uva10069 - Distinct Subsequences(大数+DP）

题目大意：给出字符串A ， B。问能够在A中找到多少子串B，可以不连续。

解题思路：dp【i】【j】 代表从i位开始的B串在从j位开始的A串中能够找到多少种。

B【i】 == A【j】 dp【i】【j】 = dp【i - 1】【j - 1】 + dp【i】【j - 1】；

B【i】 ！= A【j】 dp【i】【j】 = dp【i】【j - 1】；边界要处理一下。就是B中只有最后的那个字符来和A匹配的情况要处理一下。

10^100。要用大数。

代码：

#include <cstdio>
#include <cstring>
#include <string>

const int N = 10005;
const int M = 105;
const int base = 100000;

char s1
, s2[M];

int max (const int a, const int b) { return a > b ? a: b; }

struct bign {

int len, s[30];
bign () { memset (s, 0, sizeof (s));}
bign (int num) { *this = num;}
bign (const bign& b) { *this = b;}
bign operator = (int num);
bign operator + (const bign& b);
bign operator + (const int b);
bign operator += (const bign& b);
void DelZore ();
} dp[M]
;

void bign::DelZore () {

while (len >= 0 && s[len - 1] == 0) {
len--;
}
if (len == 0)
len = 1;
}

bign bign::operator = (int num) {

if (num == 0) {

len = 1;
s[0] = 0;
} else {

len = 0;
while (num > 0) {

s[len++] = num % base;
num = num / base;
}
}
return * this;
}

bign bign::operator + (const bign& b) {

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 % base;
g = x / base;
}
return c;
}

bign bign::operator + (const int b) {

bign b1;
b1 = b;
return *this + b1;
}

bign bign::operator += (const bign& b) {

*this = *this + b;
return *this;
}

int main () {

int t, l1, l2;
scanf ("%d%*c", &t);
while (t--) {

gets(s1);
gets(s2);
l1 = strlen (s1);
l2 = strlen (s2);

if (l2 == 0) {

printf ("0\n");
continue;
}

for (int i = 0; i < l2; i++) {
dp[i][l1] = 0;
}

for (int i = l1 - 1; i >= 0; i--) {
if (s1[i] == s2[l2 - 1])
dp[l2 - 1][i] = dp[l2 - 1][i + 1] + 1;
else
dp[l2 - 1][i] = dp[l2 - 1][i + 1];
}

for (int i = l2 - 2; i >= 0; i--)
for (int j = l1 - 1; j >= 0; j--) {

dp[i][j] = dp[i][j + 1];
if (s2[i] == s1[j])
dp[i][j] += dp[i + 1][j + 1];
}

dp[0][0].DelZore();
printf ("%d", dp[0][0].s[dp[0][0].len - 1]);
for (int i = dp[0][0].len - 2; i >= 0; i--)
printf ("%05d", dp[0][0].s[i]);
printf ("\n");
}
return 0;
}