輸入正整數n，按從小到大的順序輸出所有形如abcde/fghij=n的表達式，其中a~j恰好為數字0~9的一個排列，2n79。

題解：暴力破解枚舉fghij。

#include<iostream> #include <string> #include <string.h> #include<vector> #include<stack> #include<queue> #include<stdio.h> #include<stdlib.h> #include<iomanip> using namespace std; const int maxn = 50000; int main() { int kase=0;int a[5], b[5];int x = 0; while (cin >> x && x) {if(kase)cout<<endl;kase++;int c = 99999 / x; bool iscout = false;for (int i = 1234; i <= c; i++) {int k = i;for (int z = 0; z < 5; z++) {a[z] = k % 10; k /= 10;}bool isequal = false;for (int z = 0; z < 5; z++) {for (int j = z + 1; j < 5; j++) {if (a[z] == a[j]) {isequal = true; break;}}}if (isequal)continue;int to = i * x;if (to >= 100000)break;int tx = to;for (int z = 0; z < 5; z++) {b[z] = tx % 10; tx /= 10;}isequal = false;for (int z = 0; z < 5; z++) {for (int j = z + 1; j < 5; j++) {for (int r = 0; r < 5; r++)if (b[z] == a[r] || b[j] == a[r]) {isequal = true; break;}if (b[z] == b[j]) {isequal = true; break;}}}if (isequal)continue;/*cout << to << " / " << right << setw(5) << setfill('0') << i << " = " << x << endl;*/printf("%05d / %05d = %d\n", to, i, x);iscout = true;}if (!iscout)printf("There are no solutions for %d.\n", x);}//system("pause");return 0; }

?

總結

以上是生活随笔為你收集整理的UVA 725 Division（除法）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。