纯暴力兄弟们，九层循环，排除各个字母相等的情况。直接暴力~ ``` package com.xzy.dashoj.day03; public class xzy_05_凑算式 { public static void main(String[] args) { int count = 0; for (int a = 1; a <= 9; a++) { for (int b = 1; b <= 9; b++) { for (int c = 1; c <= 9; c++) { for (int d = 1; d <= 9; d++) { for (int e = 1; e <= 9; e++) { for (int f = 1; f <= 9; f++) { for (int g = 1; g <= 9; g++) { for (int h = 1; h <= 9; h++) { for (int i = 1; i <= 9; i++) { if (a != b && a != c && a != d && a != e && a != f && a != g && a != h && a != i && b != c && b != d && b != e && b != f && b != g && b != h && b != i && c != d && c != e && c != f && c != g && c != h && c != i && d != e && d != f && d != g && d != h && d != i && e != f && e != g && e != h && e != i && f != g && f != h && f != i && g != h && g != i && h != i) { int shang = d * 100 + e * 10 + f; int xia = g * 100 + h * 10 + i; if (a * c * xia + b * xia + shang * c == 10 * c * xia) { count++; } } } } } } } } } } } System.out.println(count); } } ```

方法一：DFS 全排列 ``` #include<bits/stdc++.h> using namespace std; const int N=15; bool used[N]; //记录该数字是否被使用过 int num[N]; //保存全排列的结果 int cnt; //计数 void dfs(int n){ if(n==10){ int A = num[1]; int B = num[2]; int C = num[3]; int DEF = num[4] * 100 + num[5] * 10 + num[6]; int GHI = num[7] * 100 + num[8] * 10 + num[9]; if(A * C * GHI + B * GHI + DEF * C == 10 * C * GHI){ cnt++; return; } } //搜索 全排列1~9 for(int i = 1;i <= 9; i++){ if(!used[i]){ used[i] = true; num[n] = i; dfs(n+1); used[i] = false; //恢复现场 } } } int main(){ dfs(1); cout<<cnt<<endl; return 0; } ``` 方法二： 利用STL 函数：next_permutation() ``` ` #include<bits/stdc++.h> using namespace std; int num[]={1,2,3,4,5,6,7,8,9}; int A,B,C,DEF,GHI; int cnt; int main(){ while(next_permutation(num,num+9)){ A=num[0]; B=num[1]; C=num[2]; DEF=num[3]*100+num[4]*10+num[5]; GHI=num[6]*100+num[7]*10+num[8]; if(A * C * GHI + B * GHI + DEF * C == 10 * C * GHI){ cnt++; } } cout<<cnt<<endl; return 0; } ``` `

import java.util.Scanner; public class Main { ``` static int[] num = new int[10]; //存储数组 static int ans = 0; static boolean[] v=new boolean[10]; //标记数组 public static void main(String[] args) { dfs(1); System.out.println(ans); } public static void dfs(int step) { //终止状态 if(step>9) { int A=num[1]; int B=num[2]; int C=num[3]; int DEF=num[4]*100+num[5]*10+num[6]; int GHI=num[7]*100+num[8]*10+num[9]; if(A*C*GHI+B*GHI+DEF*C==10*C*GHI) { ans++; return; } } //寻找新状态 for(int i=1;i<10;i++) { if(v[i]==true) { continue; //true为走过 } v[i]=true; //标记为走过 //加入队列中 num[step]=i; dfs(step+1); //回溯 v[i]=false; num[step]=0; } } ``` }

1.注意题目描述出错，数字范围是1到9而不是0到9，这东西卡我1个多小时。 2.那个DEF是一个三位数而不是DEF 因为涉及到浮点运算可能会丢失精度，所以我们把除法式子先换成乘法式子（此题中不存在分母为0的情况） include define int long long using namespace std; int a[10],ans=0; bool vis[10]={false}; bool check() { int h1=a[3]100+a[4]10+a[5],h2=a[6]100+a[7]10+a[8]; if(a[0]a[2]h2+a[1]h2+h1a[2]==10a[2]h2) return true; return false; } void dfs(int i) { if(i==9) { if(check()) ans++; return; } for(int t=1;t<10;t++) { if(!vis[t]) { a[i]=t; vis[t]=true; dfs(i+1); vis[t]=false; } } } signed main() { dfs(0); cout <<ans<< endl; return 0; }

include include using namespace std; int num; int arr[15], st[15]; void dfs(int x) { if (x == 10) { int n = arr[4] 100 + arr[5] 10 + arr[6]; int m = arr[7] 100 + arr[8] 10 + arr[9]; if (arr[1]arr[3]m + arr[2]m + n arr[3] == 10 arr[3]m) num++; return ; } for (int i = 1; i <= 9; i++) { if (st[i]) continue; st[i] = 1; arr[x] = i; dfs(x + 1); st[i] = 0; ``` } ``` } int main () { dfs(1); cout << num; return 0; }

> 你说得对，但这就是next_premutation，后面忘了 ``` #include <bits/stdc++.h> using namespace std; #define ll long long int main() { std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); double a[] = { 1,2,3,4,5,6,7,8,9 };//注意别开int，涉及到除法 ll res = 0; do { if ((a[0] + a[1] / a[2] + (a[3] * 100 + a[4] * 10 + a[5]) / (a[6] * 100 + a[7] * 10 + a[8])) == 10) ++res; } while (next_permutation(a,a+9)); cout << res; return 0; } ```

include using namespace std; int ans, n ; int a[20]; bool b[20]; void dfs(int n) { if (n == 10) { int x = a[4] 100 + a[5] 10 + a[6]; int y = a[7] 100 + a[8] 10 + a[9]; if (a[1]a[3]y + a[2]y + x a[3] == 10 a[3]y) { ans++; return; } } for (int i = 1; i < 10; i++) { if (b[i] == 1) continue; a[n] = i; b[i] = 1; dfs(n + 1); b[i] = 0; } } int main() { dfs(1); cout << ans; return 0; }

``` #include <iostream> #include <cstring> #include <algorithm> using namespace std; const int N = 15; int path[N]; bool st[N]; int n, res; void dfs(int u) { if(u > n) { for (int i = 1; i <= n; i ++ ) { int a = path[4] * 100 + path[5] * 10 + path[6]; int b = path[7] * 100 + path[8] * 10 + path[9]; if(path[1] * path[3] * b + path[2] * b + a * path[3] == 10 * path[3] * b) { res++; return; } } } for (int i = 1; i <= n; i ++ ) { if(st[i] == false) { st[i] = true; path[u] = i; dfs(u + 1); st[i] = false; } } } int main() { cin >> n; dfs(1); cout << res; return 0; } ```

``` #include<iostream> #include<algorithm> #include<cmath> #include<cstdio> #include<vector> using namespace std; vector<int> path; int used[10]; int res; void dfs(int count){ if(count>9){ int x = path[4] * 100 + path[5] * 10 + path[6]; int y = path[7] * 100 + path[8] * 10 + path[9]; // 判断是否满足条件 if (path[1] * path[3] * y + path[2] * y + x * path[3] == 10* path[3] * y){ res++; // 满足条件的排列数量加一 return; } } // 处理单层递归的逻辑 for(int i=1;i<=9;i++){ if(used[i]) continue; path.push_back(i); used[i]=1; dfs(count+1); used[i]=0; path.pop_back(); } } int main(){ path.push_back(0); dfs(1); cout<<res; return 0; } ```

```Java static int[] ans; static int[] visited; static int count; public static void main(String[] args) throws IOException { ans = new int[10]; visited = new int[10]; dfs(1); System.out.println(count); } public static void dfs(int index) { if (index == ans.length) { if ((double)ans[1] + (double) ans[2] / ans[3] + (double) (ans[4] * 100 + ans[5] * 10 + ans[6]) / (ans[7] * 100 + ans[8] * 10 + ans[9]) == 10.0) { count ++; } }else { for (int i = 1; i < 10; i++) { if (visited[i] == 0) { visited[i] = 1; ans[index] = i; dfs(index + 1); visited[i] = 0; } } } } ```