題目簡要
我們只考慮加法的蟲食算。這裏的加法是N進制加法,算式中三個數都有N位,允許有前導的0。其次,蟲子把所有的數都啃光了,我們只知道哪些數字是相同的,我們將相同的數字用相同的字母表示,不同的數字用不同的字母表示。如果這個算式是N進制的,我們就取英文字母表午的前N個大寫字母來表示這個算式中的0到N-1這N個不同的數字:但是這N個字母並不一定順序地代表0到N-1。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <algorithm>
using namespace std;
int N;
string equ[3];
int alphabet2Num[30];
int right2Left[30];//存字母序號
int used[30];
int str2Num(string s){
int num = 0,tmp = 1;
for (int i = s.length() - 1; i >= 0; --i) {
num += alphabet2Num[s[i]-'A'] * tmp;
tmp *= N;
}
return num;
}
//最後的轉換也可能耗時間
bool check(){
int num1,num2,num3;
int tmp;
num1 = str2Num(equ[0]);
num2 = str2Num(equ[1]);
num3 = str2Num(equ[2]);
tmp = num1 + num2;
return (tmp == num3);
}
//優化取模看是否能提高速度
bool check2(){
//逐位測試
for (int i = N-2; i >= 1; --i) {
if(alphabet2Num[equ[0][i] - 'A'] != -1 && alphabet2Num[equ[1][i] - 'A'] != -1 && alphabet2Num[equ[2][i] - 'A'] != -1){
if((alphabet2Num[equ[0][i] - 'A'] + alphabet2Num[equ[1][i] - 'A']) % N != alphabet2Num[equ[2][i] - 'A'] && (alphabet2Num[equ[0][i] - 'A'] + alphabet2Num[equ[1][i] - 'A'] + 1) % N != alphabet2Num[equ[2][i] - 'A'])
return true;
}
}
return false;
}
void dfs(int n){
if(n >= N ){
if(check()){
for (int i = 0; i < N; ++i) {
cout << alphabet2Num[i] << " ";
}
exit(0);
}
return;
} else {
if(n >= 3 && (alphabet2Num[equ[0][N-1] - 'A'] + alphabet2Num[equ[1][N-1] - 'A']) % N != alphabet2Num[equ[2][N-1] - 'A']){
return;
}
if((alphabet2Num[equ[0][0] - 'A'] + alphabet2Num[equ[1][0] - 'A']) >= N)
return;
if(n >= 3 && check2())
return;
for (int i = N-1; i >= 0; --i) {
if(used[i] == 0){
used[i] = 1;
alphabet2Num[right2Left[n]] = i;
dfs(n+1);
used[i] = 0;
alphabet2Num[right2Left[n]] = -1;
}
}
}
}
//200多ms的代碼
int main(){
cin >> N;
for (int i = 0; i < 3; ++i) {
cin >> equ[i];
}
int useAlphabet[30];
for (int j = 0; j < N; ++j) {
alphabet2Num[j] = -1;
useAlphabet[j] = 0;
}
int index = 0;
for (int k = N-1; k >= 0; --k) {
for (int i = 0; i < 3; ++i) {
if(useAlphabet[equ[i][k] - 'A'] == 0){
useAlphabet[equ[i][k] - 'A'] = 1;
right2Left[index++] = equ[i][k] - 'A';
}
}
}
dfs(0);
}