Uva10069-distinct subsequences (large number + dp)

Question: uva10069-distinct subsequences (large number + dp)

Returns the string a and B. Ask how many substrings B can be found in a, which can be discontinuous.

Solution: DP [I] [J] indicates the number of types of string B starting from the I-bit in string a starting from the J-bit.

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]; process the boundary. That is, only the last character in B matches.

10 ^ 100. Use a large number.

Code:

#include <cstdio>#include <cstring>#include <string>const int N = 10005;const int M = 105;const int base = 100000;char s1[N], 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][N];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;elsedp[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;}

 

Uva10069-distinct subsequences (large number + dp)