Dp + high precision uva-10069-Distinct Subsequences

Returns the number of strings A and B in string. Solution: dp + high precision. Dp [I] [j] [k] indicates the number of the first I of string B in the first j of string A, and k indicates the number of digits, each digit represents a four-digit decimal number. If save2 [I] = save1 [j] Then dp [I] [j] = dp [I] [J-1] + dp [I-1] [J-1]; handle initialization issues. If save2 [I]! = Save1 [j] dp [I] [j] = dp [I] [J-1]; high precision: Each digit represents a four-digit decimal number, if the sum of one digit and one digit times out. Code: [cpp] # include <iostream> # include <cmath> # include <cstdio> # include <cstdlib> # include <string> # include <cstring> # include <algorithm> # include <vector> # include <map> # include <stack> # include <queue> # define eps 1e-6 # define INF (1 <20) # define PI acos (-1.0) using namespace std; char save1 [11000], save2 [120]; int dp [120] [11000] [30]; // dp [I] [j] indicates the total number of one bits of Zi in Aj, which indicates 4-bit 10000 void add (int I, int j) {int i1 = I, j1 = J-1, i2 = I-1, j2 = J-1; for (int k = 1; k <= 30; k ++) {dp [I] [j] [k + 1] + = (dp [i1] [j1] [k] + dp [i2] [j2] [k]) /10000; // carry dp [I] [j] [k] + = (dp [i1] [j1] [k] + dp [i2] [j2] [k]) % 10000;} return;} int main () {int ca; scanf ("% d", & ca); while (ca --) {scanf ("% s", save1, save2); int len1 = strlen (save1), len2 = strlen (save2); memset (dp, 0, sizeof (dp); for (int j = 0; j <= len1; j ++) // pay attention to initialization, used to construct the first scenario dp [0] [j] [1] = 1; for (int I = 1; I <= len2; I ++) f Or (int j = 1; j <= len1; j ++) {if (save2 [I-1] = save1 [J-1]) {// dp [I] [j] = dp [I] [J-1] + dp [I-1] [J-1]; // You can inherit from dp [I] [J-1] or add dp [I-1] [J-1] add (I, j );} else {// dp [I] [j] = dp [I] [J-1]; for (int k = 1; k <= 30; k ++) // directly copy dp [I] [j] [k] = dp [I] [J-1] [k];} // printf ("% lld \ n ", dp [len2] [len1]); int I; for (I = 30; dp [len2] [len1] [I] = 0 & I> = 1; I --) // remove the front zero; printf ("% d", dp [len2] [len1] [I]); // process the first one without four digits. I --; for (; I> = 1; I --) printf ("% 04d", dp [len2] [len1] [I]); // note that the front zero putchar ('\ n');} return 0 ;}