Codeforces 443a Borya and Hanabi (violent)

Link: codeforces 443a Borya and Hanabi

There are several cards, each of which has two values: Color and number. Now I want to ask how many times at least to differentiate all cards, you can determine the position of a card or a digital card each time you ask.

Solution: The color and number that needs to be asked for brute-force enumeration, Which is 210. Then, the enumeration is used to determine whether it can be distinguished.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 105;int n, l[N], r[N];inline int cal (char ch) {    if (ch == ‘B‘)        return 0;    else if (ch == ‘Y‘)        return 1;    else if (ch == ‘W‘)        return 2;    else if (ch == ‘G‘)        return 3;    else if (ch == ‘R‘)        return 4;    return -1;}int bit (int i) {    return 1<<i;}int bitCount (int x) {    return x == 0 ? 0 : bitCount(x/2) + (x&1);}bool judge (int s) {    for (int i = 0; i < n; i++) {        for (int j = i + 1; j < n; j++) {            if (l[i] == l[j]) {                if (r[i] != r[j] && (s&bit(r[i]+5)) == 0 && (s&bit(r[j]+5)) == 0)                    return false;            } else {                if ((s&bit(l[i])) || (s&bit(l[j])))                    continue;                if (r[i] != r[j] && ( (s&bit(r[i]+5)) || (s&bit(r[j]+5)) ))                    continue;                return false;            }        }    }    return true;}int main () {    char str[N];    scanf("%d", &n);    for (int i = 0; i < n; i++) {        scanf("%s", str);        l[i] = cal(str[0]);        r[i] = str[1] - ‘1‘;    }    int ans = 10;    for (int i = 0; i < (1<<10); i++) {        int cnt = bitCount(i);        if (cnt >= ans)            continue;        if (judge(i))            ans = cnt;    }    printf("%d\n", ans);    return 0;}