This is back in the Wild West where everybody is fighting everybody. In particular, there are n cowboys, each with a revolver. These are rather civilized cowboys, so they have decided to take turns firing their guns until only one is left standing. Each of them has a given probability of hitting his target, and they all know each other’s probability. Furthermore, they are geniuses and always know which person to aim at in order to maximize their winning chance, so they are indeed peculiar cowboys. If there are several equally good targets, one of those will be chosen at random. Note that a cowboy’s code of ethics forces him to do his best at killing one of his opponents, even if intentionally missing would have increased his odds (yes, this can happen!) ]]>

#include #include #include #define PID 1016 #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; } ]]> You are working as a guide on a tour bus for retired people, and today you have taken your regular Nordic seniors to The Gate of Heavenly Peace. You let them have a lunch break where they could do whatever they like. Now you have to get them back to the bus, but they are all walking in random directions. You try to intersect them, and send them straight back to the bus. Minimize the time before the last person is in the bus. You will always be able to run faster than any of the tour guests, and they walk with constant speed, no matter what you tell them. The seniors walk in straight lines, and the only way of changing their direction is to give them promises of camphor candy. A senior will neither stop at nor enter the bus before given such a promise. ]]> 6 ≤ xi, yi ≤ 10⁶), speed (1 ≤ vi < 100) and direction (0 ≤ ai < 2π) of each of the tour guests. The input is terminated by a case with n = 0, which should not be processed. All floating point numbers in the input will be written in standard decimal notation, and have no more than 10 digits. ]]>

6. ]]>

kidtime ) kidtime = time + sqrt(guidx*guidx + guidy*guidy)/a[ record[i] ][2]; } if(kidtime < minimum) minimum =kidtime; }while( next_permutation( record , record+n ) ); //next_permutation printf("%.0lf\n",minimum); } return 1; } ]]> The magician shuffles a small pack of cards, holds it face down and performs the following procedure:

• The top card is moved to the bottom of the pack. The new top card is dealt face up onto the table. It is the Ace of Spades.
• Two cards are moved one at a time from the top to the bottom. The next card is dealt face up onto the table. It is the Two of Spades.
• Three cards are moved one at a time…
• This goes on until the nth and last card turns out to be the n of Spades.

This impressive trick works if the magician knows how to arrange the cards beforehand (and knows how to give a false shuffle). Your program has to determine the initial order of the cards for a given number of cards, 1 ≤ n ≤ 13. ]]>

void main() { int n,m; scanf("%d",&m); while(m--) {scanf("%d",&n); if(n==1)printf("1\n"); if(n==2)printf("2 1\n"); if(n==3)printf("3 1 2\n"); if(n==4)printf("2 1 4 3\n"); if(n==5)printf("3 1 4 5 2\n"); if(n==6)printf("4 1 6 3 2 5\n"); if(n==7)printf("5 1 3 4 2 6 7\n"); if(n==8)printf("3 1 7 5 2 6 8 4\n"); if(n==9)printf("7 1 8 6 2 9 4 5 3\n"); if(n==10)printf("9 1 8 5 2 4 7 6 3 10\n"); if(n==11)printf("5 1 6 4 2 10 11 7 3 8 9\n"); if(n==12)printf("7 1 4 9 2 11 10 8 3 6 5 12\n"); if(n==13)printf("4 1 13 11 2 10 6 7 3 5 12 9 8\n"); } } ]]> The set of cyclic rotations of a string are the strings obtained by embedding the string clockwise on a ring, with the first character following on the last, starting at any character position and moving clockwise on the ring until the character preceeding the starting character is reached. A string is a necklace if it is the lexicographically smallest among all its cyclic rotations. For instance, for the string 01011 the cyclic rotations are (10110,01101,11010,10101,01011), and furthermore 01011 is the smallest string and hence, a necklace. Any string S can be written in a unique way as a concatenation S = T1T2 . . . Tk of necklaces Ti such that Ti+1 < Ti for all i = 1, . . . , k - 1, and TiTi+1 is not a necklace for any i = 1, . . . , k - 1. This representation is called the necklace decomposition of the string S, and your task is to find it. The relation < on two strings is the lexicographical order and has the usual interpretation: A < B if A is a proper prefix of B or if A is equal to B in the first j - 1 positions but smaller in the jth position for some j. For instance, 001 < 0010 and 1101011 < 1101100. ]]>

#include int eva(char *num, int len) { int i; char sum[512]; for (i = 1 ; i < len ; i++) { memset(sum, 0, sizeof(sum)); strncpy(sum, num + i, len - i); strncpy(sum + len - i, num, i); if (strncmp(num, sum, len) > 0) return 0; } return 1; } int main(void) { int n, i, j; int len; char *p; char num[512]; scanf("%d%*c", &n); while (n--) { gets(num); p = num; while (len = strlen(p)) { for (j = len ; !eva(p, j) ; j--) ; putchar('('); for (i = 0 ; i < j ; i++) putchar(*(p + i)); putchar(')'); p += j; } putchar('\n'); } return 0; }]]>