hdu3699 A hard Aoshu Problem

正解是DFS每种字母代表的数字

其实暴力枚举就可以了

#include <cstdio>#include <ctime>#include <cstdlib>#include <cstring>#include <queue>#include <string>#include <set>#include <stack>#include <map>#include <cmath>#include <vector>#include <iostream>#include <algorithm>#include <bitset>#include <fstream>#include <sstream>using namespace std;//LOOP#define FF(i, a, b) for(int i = (a); i < (b); ++i)#define FE(i, a, b) for(int i = (a); i <= (b); ++i)#define FED(i, b, a) for(int i = (b); i>= (a); --i)#define REP(i, N) for(int i = 0; i < (N); ++i)#define CLR(A,value) memset(A,value,sizeof(A))#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)//INPUT#define RI(n) scanf("%d", &n)#define RII(n, m) scanf("%d%d", &n, &m)#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)#define RS(s) scanf("%s", s)typedef long long LL;typedef vector <int> VI;const int INF = 1000000007;const double eps = 1e-10;const int MOD = 1000000000;const int MAXN = 10050 * 30;int up[10], len1, len2, len3, num[10];bool f[10];char s1[20], s2[20], s3[20];int fun(int a, int b, int c, int d, int e){//    cout << a << b << c << d << e << endl;    LL x, y, z; int ans;    x = y = z = ans = 0;    num[0] = a, num[1] = b, num[2] = c, num[3] = d, num[4] = e;    REP(i, 5)        FF(j, i + 1, 5)            if (f[i] && f[j] && num[i] == num[j])                return 0;    REP(i, len1)        x *= 10, x += num[s1[i] - 'A'];    REP(i, len2)        y *= 10, y += num[s2[i] - 'A'];    REP(i, len3)        z *= 10, z += num[s3[i] - 'A'];    if (len1 > 1 && num[s1[0] - 'A'] == 0)  return 0;    if (len2 > 1 && num[s2[0] - 'A'] == 0)  return 0;    if (len3 > 1 && num[s3[0] - 'A'] == 0)  return 0;    if (x + y == z) ans++;    if (x - y == z) ans++;    if (x * y == z) ans++;    if (y && x == z * y) ans++;    return ans;}int solve(){    int ans = 0;    REP(a, up[0])        REP(b, up[1])            REP(c, up[2])                REP(d, up[3])                    REP(e, up[4])                        ans += fun(a, b, c, d, e);    return ans;}int main(){    int T;    RI(T);    while (T--)    {        RS(s1), RS(s2), RS(s3);        CLR(f, 0);        len1 = strlen(s1), len2 = strlen(s2), len3 = strlen(s3);        REP(i, strlen(s1))            f[s1[i] - 'A'] = 1;        REP(i, strlen(s2))            f[s2[i] - 'A'] = 1;        REP(i, strlen(s3))            f[s3[i] - 'A'] = 1;        REP(i, 5)            if (f[i])   up[i] = 10;            else    up[i] = 1;        cout << solve() << endl;    }}