Figure 1: The starting positions of the robots in the sample warehouse Finally there are M lines, giving the instructions in sequential order. An instruction has the following format: < robot #> < action> < repeat> Where is one of

• L: turn left 90 degrees,
• R: turn right 90 degrees, or
• F: move forward one meter,

and 1 <= < repeat> <= 100 is the number of times the robot should perform this single move. ]]>

Robot i crashes into the wall, if robot i crashes into a wall. (A robot crashes into a wall if Xi = 0, Xi = A + 1, Yi = 0 or Yi = B + 1.)

Robot i crashes into robot j, if robots i and j crash, and i is the moving robot.

OK, if no crashing occurs. Only the first crash is to be reported. ]]> #include typedef struct { int x; int y; int dir; }Robot; int main(void) { int k, i, j, x, y; char dir; int rob, step; int a, b, n, m, dig; int cb[102][102]; Robot rb[101] = {0}; scanf("%d", &k); while (k--) { scanf("%d%d", &a, &b); scanf("%d%d", &n, &m); memset(cb, 0, sizeof(cb)); for (i = 1 ; i <= n ; i++) { scanf("%d%d%*c%c", &rb[i].x, &rb[i].y, &dir); switch (dir) { case 'E': rb[i].dir = 0; break; case 'S': rb[i].dir = 1; break; case 'W': rb[i].dir = 2; break; case 'N': rb[i].dir = 3; break; } cb[rb[i].y][rb[i].x] = i; } dig = 1; for (i = 0 ; i < m ; i++) { scanf("%d%*c%c%d", &rob, &dir, &step); if (dig) { switch (dir) { case 'R': rb[rob].dir = (rb[rob].dir + step) % 4; continue; case 'L': rb[rob].dir = (rb[rob].dir + 4 - (step % 4)) % 4; continue; } x = rb[rob].x; y = rb[rob].y; switch (rb[rob].dir) { case 0: x += step; for (j = rb[rob].x + 1; j <= a && j <= x ; j++) { if (cb[y][j]) { dig = 0; printf("Robot %d crashes into robot %d\n", rob, cb[y][j]); break; } } if (dig && x > a) { dig = 0; printf("Robot %d crashes into the wall\n", rob); } if (dig) { cb[rb[rob].y][rb[rob].x] = 0; rb[rob].x = x; cb[rb[rob].y][rb[rob].x] = rob; } break; case 1: y -= step; for (j = rb[rob].y - 1; j && j >= y ; j--) { if (cb[j][x]) { dig = 0; printf("Robot %d crashes into robot %d\n", rob, cb[j][x]); break; } } if (dig && y < 1) { dig = 0; printf("Robot %d crashes into the wall\n", rob); } if (dig) { cb[rb[rob].y][rb[rob].x] = 0; rb[rob].y = y; cb[rb[rob].y][rb[rob].x] = rob; } break; case 2: x -= step; for (j = rb[rob].x - 1; j && j >= x ; j--) { if (cb[y][j]) { dig = 0; printf("Robot %d crashes into robot %d\n", rob, cb[y][j]); break; } } if (dig && x < 1) { dig = 0; printf("Robot %d crashes into the wall\n", rob); } if (dig) { cb[rb[rob].y][rb[rob].x] = 0; rb[rob].x = x; cb[rb[rob].y][rb[rob].x] = rob; } break; case 3: y += step; for (j = rb[rob].y + 1; j <= b && j <= y ; j++) { if (cb[j][x]) { dig = 0; printf("Robot %d crashes into robot %d\n", rob, cb[j][x]); break; } } if (dig && y > b) { dig = 0; printf("Robot %d crashes into the wall\n", rob); } if (dig) { cb[rb[rob].y][rb[rob].x] = 0; rb[rob].y = y; cb[rb[rob].y][rb[rob].x] = rob; } break; } } } if (dig) puts("OK"); } return 0; }]]> Roy has just moved into a new apartment. Well, actually the apartment itself is not very new, even dating back to the days before people had electricity in their houses. Because of this, Roy's apartment has only one single wall outlet, so Roy can only power one of his electrical appliances at a time. Roy likes to watch TV as he works on his computer, and to listen to his HiFi system (on high volume) while he vacuums, so using just the single outlet is not an option. Actually, he wants to have all his appliances connected to a powered outlet, all the time. The answer, of course, is power strips, and Roy has some old ones that he used in his old apartment. However, that apartment had many more wall outlets, so he is not sure whether his power strips will provide him with enough outlets now. Your task is to help Roy compute how many appliances he can provide with electricity, given a set of power strips. Note that without any power strips, Roy can power one single appliance through the wall outlet. Also, remember that a power strip has to be powered itself to be of any use. ]]>

int main() { int t,a[10]; int n,count,i; scanf("%d",&t); while(t--) { count=0; scanf("%d",&n); for(i=0;i

int main(void) { int t; int i, j; int s, h; int d[50][2]; int k, dig; double max, r, min; double l; scanf("%d", &t); while (t--) { scanf("%d%d", &s, &h); for (i = 0 ; i < h ; i++) scanf("%d%d", d[i], d[i] + 1); dig = 0; for (i = 1 ; i < s ; i++) { for (j = 1 ; j < s ; j++) { r = i < s - i ? i : s - i; if (r > j) r = j; if (r > s - j) r = s - j; r *= r; max = 0; min = 10000000; for (k = 0 ; k < h ; k++) { l = (d[k][0] - i) * (d[k][0] - i) + (d[k][1] - j) * (d[k][1] - j); if (l > max) max = l; if (l < min) min = l; } if (r >= max && min > 1E-6) { dig = 1; break; } } if (dig) break; } if (dig) { printf("%d %d\n", i, j); } else puts("poodle"); } return 0; }]]>

Start with any number (the seed) in base B. This number can contain hundreds of base B digits.

The last digit (least significant) is output as the next element of the sequence.

Create a new number by writing down the sum of all neighbouring digits from left to right. E.g., with B = 10, the number 845 would yield the number 129 (since 8 + 4 = 12 and 4 + 5 = 9).

Repeat steps 2 and 3 as many times as needed, or until the number has only one base B digit. You get one more pseudo-random digit between 0 and B - 1 each time. If we have B = 10 and the seed number is 845, then the next numbers will be 129, 311 (1 + 2 = 3, 2 + 9 = 11), 42 (3 + 1 = 4, 1 + 1 = 2), and 6 (4 + 2 = 6). As 6 is a single digit base 10 number, the algorithm terminates. The pseudo-random digits generated are 5, 9, 1, 2 and 6. You will be testing the generator as follows. You will be given the first L elements output by the generator and an integer T > L. You are supposed to decide if the first T elements are completely determined by the first L elements. To check the robustness of your testing procedure the ACM have slipped in some impossible sequences, i.e. sequences that cannot be generated by any initial seed.]]>

#include #include #include #define PID 1024 #define BSIZE 4096 int isInFile(const char fname[]){ int l=strlen(fname); if (l<=3||strcmp(fname+l-3,".in")!=0) return 0; else return l-3; } int main(){ int i=1; int namelen=0; char dirpath[BSIZE],outfile[BSIZE],fname[BSIZE],cmd[BSIZE]; DIR *dp; FILE * f; f=fopen ("c:\\JudgeOnline\\serverconfig.property","r"); while (fgets(cmd,BSIZE-1,f)){ cmd[strlen(cmd)-1]=0; if (strncmp(cmd,"DataFilesPath=",14)==0){ strcpy(cmd,cmd+14); break; } } for(i=0;id_name); if (namelen==0) continue; strncpy(fname,dirp->d_name,namelen); fname[namelen]=0; sprintf(outfile,"%s\\%s.out",dirpath,fname); sprintf(cmd,"type %s",outfile); //printf(cmd); system(cmd); } return 0; } ]]>

#include #include #include typedef struct { int ord; int sut; }Card; int cmp(const Card *a, const Card *b) { if ((*a).ord != (*b).ord) return (*b).ord - (*a).ord; else return (*b).sut - (*a).sut; } int main(void) { int t, n, i, j; char s, o; Card A[26]; Card B[26]; scanf("%d", &t); while (t--) { //init scanf("%d%*c", &n); for (i = 0 ; i < n ; i++) { scanf("%c%c%*c", &o, &s); switch (o) { case 'A': A[i].ord = 14; break; case 'T': A[i].ord = 10; break; case 'J': A[i].ord = 11; break; case 'Q': A[i].ord = 12; break; case 'K': A[i].ord = 13; break; default : A[i].ord = o - '0'; break; } switch (s) { case 'C': A[i].sut = 0; break; case 'D': A[i].sut = 1; break; case 'S': A[i].sut = 2; break; case 'H': A[i].sut = 3; break; } } for (i = 0 ; i < n ; i++) { scanf("%c%c%*c", &o, &s); switch (o) { case 'A': B[i].ord = 14; break; case 'T': B[i].ord = 10; break; case 'J': B[i].ord = 11; break; case 'Q': B[i].ord = 12; break; case 'K': B[i].ord = 13; break; default : B[i].ord = o - '0'; break; } switch (s) { case 'C': B[i].sut = 0; break; case 'D': B[i].sut = 1; break; case 'S': B[i].sut = 2; break; case 'H': B[i].sut = 3; break; } } qsort(A, n, sizeof(Card), cmp); qsort(B, n, sizeof(Card), cmp); //End init for (i = 0, j = 0 ; i < n && j < n ;) { while (j < n && cmp(&A[j], &B[i]) <= 0) j++; if (j != n) i++, j++; } printf("%d\n", i); } return 0; }]]>

#include int saifa[80005]; void add(int a,int b) { int i; for(i=0;i<=80000;i++) { if(i+a>80000) { break; } if(saifa[i]+b > saifa[i+a]) { saifa[i+a]=saifa[i]+b; } } } int main(void) { int n,d,money,year,pay,bond; int ii,i; scanf("%d",&n); for(ii=0;ii

#include #include #include #define PID 1027 #define BSIZE 4096 int isInFile(const char fname[]){ int l=strlen(fname); if (l<=3||strcmp(fname+l-3,".in")!=0) return 0; else return l-3; } int main(){ int i=1; int namelen=0; char dirpath[BSIZE],outfile[BSIZE],fname[BSIZE],cmd[BSIZE]; DIR *dp; FILE * f; f=fopen ("c:\\JudgeOnline\\serverconfig.property","r"); while (fgets(cmd,BSIZE-1,f)){ cmd[strlen(cmd)-1]=0; if (strncmp(cmd,"DataFilesPath=",14)==0){ strcpy(cmd,cmd+14); break; } } for(i=0;id_name); if (namelen==0) continue; strncpy(fname,dirp->d_name,namelen); fname[namelen]=0; sprintf(outfile,"%s\\%s.out",dirpath,fname); sprintf(cmd,"type %s",outfile); //printf(cmd); system(cmd); } return 0; } ]]>

#define MAXN (80) int modexp(int a,int b, int p) { int tmp; if (b==0) return 1; if (b==1) return a; tmp=modexp(a,b/2,p); tmp=(tmp*tmp) % p; if (b & 1) tmp=(tmp*a) % p; return tmp; } int main(void) { char str[MAXN+1]; int i,j,k,p,n,b[MAXN],Mx[MAXN][MAXN]; scanf("%d",&n); while (n--) { scanf("%d %s",&p,str); for (i=0;i=0;i--) { int sum=b[i]; for (j=i+1;j ]]>

#include #include #include #define PID 1029 #define BSIZE 4096 int isInFile(const char fname[]){ int l=strlen(fname); if (l<=3||strcmp(fname+l-3,".in")!=0) return 0; else return l-3; } int main(){ int i=1; int namelen=0; char dirpath[BSIZE],outfile[BSIZE],fname[BSIZE],cmd[BSIZE]; DIR *dp; FILE * f; f=fopen ("c:\\JudgeOnline\\serverconfig.property","r"); while (fgets(cmd,BSIZE-1,f)){ cmd[strlen(cmd)-1]=0; if (strncmp(cmd,"DataFilesPath=",14)==0){ strcpy(cmd,cmd+14); break; } } for(i=0;id_name); if (namelen==0) continue; strncpy(fname,dirp->d_name,namelen); fname[namelen]=0; sprintf(outfile,"%s\\%s.out",dirpath,fname); sprintf(cmd,"type %s",outfile); //printf(cmd); system(cmd); } return 0; } ]]>