B. The least round way time limit per test5 seconds memory limit per test64 megabytes inputstandard input outputstandard output There is a square matrix n × n, consisting of non-negative integer numbers. You should find such a way on it that starts in the upper left cell of the matrix; each following cell is to the right or down from the current cell; the way ends in the bottom right cell. Moreover, if we multiply together all the numbers along the way, the result should be the least "round". In other words, it should end in the least possible number of zeros. Input The first line contains an integer number n (2 ≤ n ≤ 1000), n is the size of the matrix. Then follow n lines containing the matrix elements (non-negative integer numbers not exceeding 109). Output In the first line print the least number of trailing zeros. In the second line print the correspondent way itself. Sample test(s) input 3 1 2 3 4 5 6 7 8 9 output 0 DDRR
#include <stdio.h> #include <string.h> #include <algorithm> using namespace std; int a[1003][1003]; char path[2003]; void judge(int &x2, int &x5, int n) { x2 = x5 = 0; while(n % 5 == 0 && n) { n /= 5; x5++; } while(n % 2 == 0 && n) { n /= 2; x2++; } } struct Node { int Susake2; int Susake5; int px, py; }; Node Susake[1003][1003], dp1[1003][1003], dp2[1003][1003]; int main(int argc, char *argv[]) { int n, x2, x5, Min1, Min2, k1, k2, flag, ex, ey, tx, ty, si, sj, God; memset(a, 0, sizeof(a)); memset(Susake, 0, sizeof(Susake)); memset(path, 0, sizeof(path)); scanf("%d", &n); flag = 0; for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) { scanf("%d", &a[i][j]); if(a[i][j] == 0 && flag == 0) { flag = 1; si = i; sj = j; } judge(x2, x5, a[i][j]); Susake[i][j].Susake2 = x2; Susake[i][j].Susake5 = x5; } //bround dp1[0][0].Susake2 = 0; dp1[1][0].Susake2 = 0; dp1[0][1].Susake2 = 0; dp2[0][0].Susake5 = 0; dp2[1][0].Susake5 = 0; dp2[0][1].Susake5 = 0; for(int i = 2; i <= n; i++) { dp1[i][0].Susake2 = 1000000009; dp1[0][i].Susake2 = 1000000009; dp2[i][0].Susake5 = 1000000009; dp2[0][i].Susake5 = 1000000009; } //according to 2 dp for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) { dp1[i][j].Susake2 = min(dp1[i - 1][j].Susake2, dp1[i][j - 1].Susake2) + Susake[i][j].Susake2; //Save path if(dp1[i - 1][j].Susake2 < dp1[i][j - 1].Susake2) { dp1[i][j].px = i - 1; dp1[i][j].py = j; } if(dp1[i - 1][j].Susake2 >= dp1[i][j - 1].Susake2) { dp1[i][j].px = i; dp1[i][j].py = j - 1; } } Min1 = dp1[n][n].Susake2; //according to 5 dp for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) { dp2[i][j].Susake5 = min(dp2[i - 1][j].Susake5, dp2[i][j - 1].Susake5) + Susake[i][j].Susake5; //Save path if(dp2[i - 1][j].Susake5 < dp2[i][j - 1].Susake5) { dp2[i][j].px = i - 1; dp2[i][j].py = j; } if(dp2[i - 1][j].Susake5 >= dp2[i][j - 1].Susake5) { dp2[i][j].px = i; dp2[i][j].py = j - 1; } } Min2 = dp2[n][n].Susake5; //print the path dp1[1][1].px = 1; dp1[1][1].py = 1; dp2[1][1].px = 1; dp2[1][1].py = 1; God = 0; if(min(Min1, Min2) > 1 && flag) { printf("%d\n", 1); for(int i = 2; i <= si; i++) printf("D"); for(int i = 2; i <= sj; i++) printf("R"); for(int i = si + 1; i <= n; i++) printf("D"); for(int i = sj + 1; i <= n; i++) printf("R"); printf("\n"); } else { if(Min1 <= Min2) { k1 = 0; tx = ty = n; ex = tx; ey = ty; tx = dp1[tx][ty].px; ty = dp1[tx][ty].py; if(a[ex][ey] == 0) God = 1; if(a[tx][ty] == 0) God = 1; if(tx < ex) { k1++; path[k1] = 'D'; } if(ty < ey) { k1++; path[k1] = 'R'; } ex = tx; ey = ty; while(tx != 1 || ty != 1) { tx = dp1[ex][ey].px; ty = dp1[ex][ey].py; if(a[tx][ty] == 0) God = 1; if(tx < ex) { k1++; path[k1] = 'D'; } if(ty < ey) { k1++; path[k1] = 'R'; } ex = tx; ey = ty; } if(God == 1) printf("1\n"); else printf("%d\n", min(Min1, Min2)); for(int i = k1; i >= 1; i--) printf("%c", path[i]); printf("\n"); } else { k2 = 0; tx = ty = n; ex = tx; ey = ty; tx = dp2[tx][ty].px; ty = dp2[tx][ty].py; if(a[ex][ey] == 0) God = 1; if(a[tx][ty] == 0) God = 1; if(tx < ex) { k2++; path[k2] = 'D'; } if(ty < ey) { k2++; path[k2] = 'R'; } ex = tx; ey = ty; while(tx != 1 || ty != 1) { tx = dp2[ex][ey].px; ty = dp2[ex][ey].py; if(a[tx][ty] == 0) God = 1; if(tx < ex) { k2++; path[k2] = 'D'; } if(ty < ey) { k2++; path[k2] = 'R'; } ex = tx; ey = ty; } if(God == 1) printf("1\n"); else printf("%d\n", min(Min1, Min2)); for(int i = k2; i >= 1; i--) printf("%c", path[i]); printf("\n"); } } return 0; }