CF 2B The least round way (DP)

Question: an n * n matrix, which can only be down or right at a time. From top left to bottom right, all the numbers in a path are multiplied, judge at least a few 0 http://codeforces.com/problemset/problem/2/ B at the end of the group to see the cross sister recommended, it is done, you deserve it! The end of the multiplication is 0, indicating 2*5. The final product can be expressed as 2 ^ a * 5 ^ B * other. Then the number of the last 0 is min (a, B) dp [I] [j] [0] indicates the minimum number of record factors 2 dp [I] [j] [1] indicates the minimum number of record factors 5, but there is one trick, that is, the number in the matrix can be zero, and the product is 0. Therefore, take 0 as 10 to run DP once. If the result is 0, there is a path that does not go through 0, and there is no 0 at the end. If the result is greater than 1, you can select the path that goes through 0, the answer is 1 [cpp] # include <iostream> # include <cstdio> # include <map> # include <cstring> # include <cmath> # include <vector> # include <algorithm> # include <set> # include <string> # include <queue> # define inf 1000000005 # define M 40 # define N 10005 # define maxn 300005 # define eps 1e-8 # define zero () fabs (a) <eps # define Min (a, B) (a) <(B )? (A) :( B) # define Max (a, B) (a)> (B )? (A) :( B) # define pb (a) push_back (a) # define mp (a, B) make_pair (a, B) # define mem (a, B) memset (a, B, sizeof ()) # define LL long # define MOD 1000000009 # define lson step <1 # define rson step <1 | 1 # define sqr (a) * ()) # define Key_value ch [ch [root] [1] [0] # define test puts ("OK"); # define pi acos (-1.0) # define lowbit (x) (-(x) & (x) # define HASH1 1331 # define HASH2 10001 # pragma comment (linker, "/ST ACK: 10241000000,1024000000 ") using namespace std; // dp [I] [j] [k] [l] indicates arrival (I, j), l = 0 indicates two, l = 1 indicates five int dp [1005] [1005] [2] [2]; int n, a [1005] [1005]; pair <int, pair <int, int> pre [1005] [1005] [2]; int way [2] [2] = {,}; int get (int num, int fac) {int ret = 0; if (! Num) return 1; while (num & (num % fac) = 0) {ret ++; num/= fac;} return ret;} int main () {// freopen ("input.txt", "r", stdin); while (scanf ("% d", & n )! = EOF) {int zx =-1, zy =-1; for (int I = 1; I <= n; I ++) for (int j = 1; j <= n; j ++) {scanf ("% d", & a [I] [j]); if (a [I] [j] = 0) zx = I, zy = j;} mem (dp,-1); for (int j = 0; j <2; j ++) for (int k = 0; k <2; k ++) dp [0] [1] [j] [k] = dp [1] [0] [j] [k] = 0; for (int I = 1; I <= n; I ++) {for (int j = 1; j <= n; j ++) {int two = get (a [I] [j], 2), five = get (a [I] [j], 5); for (int r = 0; r <2; r ++) {int x = I-way [r] [0], y = j-way [r] [1]; for (int k = 0; k <2; k ++) {if (dp [x] [y] [k] [0] =-1 | dp [x] [Y] [k] [1] =-1) continue; int now_two = dp [x] [y] [k] [0] + two, now_five = dp [x] [y] [k] [1] + five; if (dp [I] [j] [k] [0] =-1) {dp [I] [j] [k] [0] = now_two; dp [I] [j] [k] [1] = now_five; pre [I] [j] [k] = mp (k, mp (x, y ));} else if (k & now_two <dp [I] [j] [k] [0]) {dp [I] [j] [k] [0] = now_two; dp [I] [j] [k] [1] = now_five; pre [I] [j] [k] = mp (k, mp (x, y ));} else if (! K & now_five <dp [I] [j] [k] [1]) {dp [I] [j] [k] [0] = now_two; dp [I] [j] [k] [1] = now_five; pre [I] [j] [k] = mp (k, mp (x, y) ;}}}} int ans = inf, k; for (int I = 0; I <2; I ++) {if (dp [n] [n] [I] [0] =-1) continue; int tmp = min (dp [n] [n] [I] [0], dp [n] [n] [I] [1]); if (tmp <ans) {ans = tmp; k = I ;}} if (ans> 1 & zx! =-1) {printf ("1 \ n"); for (int I = 1; I <zx; I ++) putchar ('D '); for (int j = 1; j <zy; j ++) putchar ('R'); for (int I = zx; I <n; I ++) putchar ('D'); for (int j = zy; j <n; j ++) putchar ('R'); continue;} string ret = ""; int x = n, y = n; while (x! = 1 | y! = 1) {int xx = pre [x] [y] [k]. second. first, yy = pre [x] [y] [k]. second. second; if (xx = x) ret + = 'R'; else if (yy = y) ret + = 'D '; k = pre [x] [y] [k]. first; x = xx; y = yy;} reverse (ret. begin (), ret. end (); cout <ans <endl <ret <endl;} return 0 ;} # include <iostream> # include <cstdio> # include <map> # include <cstring> # include <cmath> # include <vector> # include <algorithm> # include <set> # include <string> # include <queue> # define inf 1000000005 # define M 40 # define N 10005 # define maxn 300005 # define eps 1e-8 # define zero (a) fabs (a) <eps # define Min (a, B) (a) <(B )? (A) :( B) # define Max (a, B) (a)> (B )? (A) :( B) # define pb (a) push_back (a) # define mp (a, B) make_pair (a, B) # define mem (a, B) memset (a, B, sizeof ()) # define LL long # define MOD 1000000009 # define lson step <1 # define rson step <1 | 1 # define sqr (a) * ()) # define Key_value ch [ch [root] [1] [0] # define test puts ("OK"); # define pi acos (-1.0) # define lowbit (x) (-(x) & (x) # define HASH1 1331 # define HASH2 10001 # pragma comment (linker, "/STACK: 102 40010000,1024000000 ") using namespace std; // dp [I] [j] [k] [l] indicates arrival (I, j), l = 0 indicates two, l = 1 indicates fiveint dp [1005] [1005] [2] [2]; int n, a [1005] [1005]; pair <int, pair <int, int> pre [1005] [1005] [2]; int way [2] [2] = {,}; int get (int num, int fac) {int ret = 0; if (! Num) return 1; while (num & (num % fac) = 0) {ret ++; num/= fac;} return ret;} int main () {// freopen ("input.txt", "r", stdin); while (scanf ("% d", & n )! = EOF) {int zx =-1, zy =-1; for (int I = 1; I <= n; I ++) for (int j = 1; j <= n; j ++) {scanf ("% d", & a [I] [j]); if (a [I] [j] = 0) zx = I, zy = j;} mem (dp,-1); for (int j = 0; j <2; j ++) for (int k = 0; k <2; k ++) dp [0] [1] [j] [k] = dp [1] [0] [j] [k] = 0; for (int I = 1; I <= n; I ++) {for (int j = 1; j <= n; j ++) {int two = get (a [I] [j], 2), five = get (a [I] [j], 5); for (int r = 0; r <2; r ++) {int x = I-way [r] [0], y = j-way [r] [1]; for (int k = 0; k <2; k ++) {if (dp [x] [y] [k] [0] =-1 | dp [x] [Y] [k] [1] =-1) continue; int now_two = dp [x] [y] [k] [0] + two, now_five = dp [x] [y] [k] [1] + five; if (dp [I] [j] [k] [0] =-1) {dp [I] [j] [k] [0] = now_two; dp [I] [j] [k] [1] = now_five; pre [I] [j] [k] = mp (k, mp (x, y ));} else if (k & now_two <dp [I] [j] [k] [0]) {dp [I] [j] [k] [0] = now_two; dp [I] [j] [k] [1] = now_five; pre [I] [j] [k] = mp (k, mp (x, y ));} else if (! K & now_five <dp [I] [j] [k] [1]) {dp [I] [j] [k] [0] = now_two; dp [I] [j] [k] [1] = now_five; pre [I] [j] [k] = mp (k, mp (x, y) ;}}}} int ans = inf, k; for (int I = 0; I <2; I ++) {if (dp [n] [n] [I] [0] =-1) continue; int tmp = min (dp [n] [n] [I] [0], dp [n] [n] [I] [1]); if (tmp <ans) {ans = tmp; k = I ;}} if (ans> 1 & zx! =-1) {printf ("1 \ n"); for (int I = 1; I <zx; I ++) putchar ('D '); for (int j = 1; j <zy; j ++) putchar ('R'); for (int I = zx; I <n; I ++) putchar ('D'); for (int j = zy; j <n; j ++) putchar ('R'); continue;} string ret = ""; int x = n, y = n; while (x! = 1 | y! = 1) {int xx = pre [x] [y] [k]. second. first, yy = pre [x] [y] [k]. second. second; if (xx = x) ret + = 'R'; else if (yy = y) ret + = 'D '; k = pre [x] [y] [k]. first; x = xx; y = yy;} reverse (ret. begin (), ret. end (); cout <ans <endl <ret <endl;} return 0 ;}