3 August
P1013 进制位
结论:加法必为 \(n-1\) 进制;\({(n-1)}^1\) 位必为数字 1;\(0+0=0\)。
模拟、字符串。
#include <cstdio>
#include <map>
#include <cstring>
using namespace std;
int n, num[11], id[11], tab[11][11];
map<char, int> p;
map<int, char> q;
char s[3];
bool check() {
for (int i=1; i<n; ++i) for (int j=1; j<n; ++j)
if (tab[i][j]!=id[(num[i]+num[j])%(n-1)]) return false;
return true;
}
int main() {
memset(num, -1, sizeof num);
scanf("%d", &n); scanf("%s", s);
for (int i=1; i<n; ++i) {
scanf("%s", s); p[s[0]]=i, q[i]=s[0];
}
for (int i=1; i<n; ++i) {
scanf("%s", s); int k=p[s[0]];
for (int j=1; j<n; ++j) {
scanf("%s", s);
if (s[1]) num[id[1]=p[s[0]]]=1, tab[k][j]=p[s[1]];
else tab[k][j]=p[s[0]];
}
}
for (int i=1; i<n; ++i) if (tab[i][i]==i) num[id[0]=i]=0;
int k=1;
while (tab[id[1]][id[k]]!=id[0]) {
num[id[k+1]=tab[id[1]][id[k]]]=k+1; ++k;
if (k>n-1) break;
}
if (check()) {
for (int i=1; i<n; ++i) printf("%c=%d ", q[i], num[i]);
printf("\n%d\n", n-1);
} else printf("ERROR!\n");
return 0;
}
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