Description
The "Gold Bar"bank received information from reliable sources that in their last group of N coins exactly one coin is false and differs in weight from other coins (while all other coins are equal in weight). After the economic crisis they have only a simple balance available (like one in the picture). Using this balance, one is able to determine if the weight of objects in the left pan is less than, greater than, or equal to the weight of objects in the right pan. In order to detect the false coin the bank employees numbered all coins by the integers from 1 to N, thus assigning each coin a unique integer identifier. After that they began to weight various groups of coins by placing equal numbers of coins in the left pan and in the right pan. The identifiers of coins and the results of the weightings were carefully recorded. You are to write a program that will help the bank employees to determine the identifier of the false coin using the results of these weightings.Input
The first line of the input file contains two integers N and K, separated by spaces, where N is the number of coins (2<=N<=1000 ) and K is the number of weightings fulfilled (1<=K<=100). The following 2K lines describe all weightings. Two consecutive lines describe each weighting. The first of them starts with a number Pi (1<=Pi<=N/2), representing the number of coins placed in the left and in the right pans, followed by Pi identifiers of coins placed in the left pan and Pi identifiers of coins placed in the right pan. All numbers are separated by spaces. The second line contains one of the following characters: '<', '>', or '='. It represents the result of the weighting: '<' means that the weight of coins in the left pan is less than the weight of coins in the right pan, '>' means that the weight of coins in the left pan is greater than the weight of coins in the right pan, '=' means that the weight of coins in the left pan is equal to the weight of coins in the right pan.Output
Write to the output file the identifier of the false coin or 0, if it cannot be found by the results of the given weightings.Sample Input
5 3 2 1 2 3 4 < 1 1 4 = 1 2 5 =Sample Output
3Source
Northeastern Europe 1998 /* 有几个规律 1.如果出现等号，那么等号两边的金子都不是假的 2.如果出现不等号，那么一定存在假币。不等号外的那些金币必须都是假的 3.假币出现的次数等于不等号出现的次数 4.假币必须出现在不等号的同一端 我们必须满足以上四个条件 */ #include<cstdio> #include<cstring> int a[1010]; int b[1010]; int f[1010]; bool ff[1010]; int cal(int x) { if(x<0) return -x; return x; } int main() { int n,k; while(scanf("%d%d",&n,&k)!=EOF) { char s[10]; int m,num=0; memset(f,0,sizeof(f)); memset(ff,1,sizeof(ff)); for(int i=1;i<=k;i++) { scanf("%d",&m); for(int j=1;j<=m;j++) scanf("%d",&a[j]); for(int j=1;j<=m;j++) scanf("%d",&b[j]); scanf("%s",s); if(s[0]=='=') { for(int j=1;j<=m;j++) { f[a[j]]=0; f[b[j]]=0;//保证1 ff[a[j]]=0; ff[b[j]]=0; } } else { num++;//不等号出现的次数 if(s[0]=='<') { for(int j=1;j<=m;j++) f[a[j]]--,f[b[j]]++;//保证4 } else { for(int j=1;j<=m;j++) f[a[j]]++,f[b[j]]--;//保证4 } } } int mm=0; int ca=1; for(int i=1;i<=n;i++) if(ff[i]&&cal(f[i])==num) { mm++; ca=i; } if(mm==1) printf("%d/n",ca); else printf("0/n"); } return 0; }