字符串模拟+并查集
建立两个并查集分别存放每个变量的每一位数的祖先，一个是1一个是2
考虑每个字母的每一位的数都是唯一的，先模拟，记录每一个变量的每一位。
一一映射到方程中去，最后将两个方程进行一一比较，然后合并并查集。中间判断是否出现一位既是1又是2的情况
最后统计自由元的个数cnt，高精求解2^cnt
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define int long long
#define N 100100
using namespace std;
int k, tot, len, len1, len2, tot1, tot2;
int x[N], y[N], sum[N], fa[N], jin[N]; //sum[i]表示i所代表的字母的开始位置
string s1, s2;
int find(int a)
{
if (a == fa[a]) return a;
return fa[a] = find(fa[a]);
}
signed main()
{
scanf("%lld", &k);
sum[1] = len = 2;
for (int i = 2; i <= k + 1; i++)
scanf("%lld", &len), sum[i] = sum[i - 1] + len, tot += len;
cin >> s1 >> s2;
int len1 = s1.size(), len2 = s2.size();
for (int i = 0; i < len1; i++)
{
if (s1[i] >= 'a' && s1[i] <= 'z')
{
int now = s1[i] - 'a' + 1;
for (int j = sum[now]; j < sum[now + 1]; j++)
x[++tot1] = j;
}
else x[++tot1] = s1[i] - '0';
}
for (int i = 0; i < len2; i++)
{
if (s2[i] >= 'a' && s2[i] <= 'z')
{
int now = s2[i] - 'a' + 1;
for (int j = sum[now]; j < sum[now + 1]; j++)
y[++tot2] = j;
}
else y[++tot2] = s2[i] - '0';
}
if (tot1 != tot2)
printf("0"), exit(0);
for (int i = 1; i <= tot1 * 4; i++)
fa[i] = i;
for (int i = 1; i <= tot1; i++)
{
int a = x[i], b = y[i];
if (a + b == 1) //如果a，b不相等，就直接输出不行
printf("%lld
", 0), exit(0);
int da = find(a), db = find(b);
if (da + db == 1)
printf("0"), exit(0);
if (da != db) //已经排除了a，b等于1或0的情况
fa[da] = db, tot--;
}
int big[100010]= {1}, top=1;
for(int i=tot; i>=1; i--)
{
for(int i=0; i<top; ++i)big[i]<<=1;
for(int i=0; i<top; ++i)if(big[i]>=10)
{
big[i+1]+=big[i]/10,big[i]%=10;
}
for(; big[top]; ++top)
{
big[top+1]+=big[top]/10,big[top]%=10;
}
}
for(int i=top-1; i>=0; --i) printf("%lld", big[i]);
return 0;
}
/*
5
4 2 4 4 2
1bad1
acbe
*/