随手练——洛谷-P1008 / P1618 三连击（暴力搜索）

1.普通版

 

第一眼看到这个题，我脑海里就是，“我们是不是在哪里见过~”，去年大一刚学C语言的时候写过一个类似的题目，写了九重循环。。。。就像这样（在洛谷题解里看到一位兄台写的。。。。超长警告，慎重点开）

#include <stdio.h>
#include <cstdlib>
int main()
{int i[9];for (i[0] = 1; i[0] <= 9; i[0]++){for (i[1] = 1; i[1] <= 9; i[1]++){int p1=0;if (i[1] == i[0]) p1 = 1;if (p1 != 1) {for (i[2] = 1; i[2] <= 9; i[2]++){int p2=0;for (int j2 = 0; j2 < 2; j2++) if (i[2] == i[j2]) p2 = 2;if (p2 != 2) {for (i[3] = 1; i[3] <= 9; i[3]++){int p3=0;for (int j3 = 0; j3 < 3; j3++) if (i[3] == i[j3]) p3 = 3;if (p3 != 3) {for (i[4] = 1; i[4] <= 9; i[4]++){int p4=0;for (int j4 = 0; j4 < 4; j4++) if (i[4] == i[j4]) p4 = 4;if (p4 != 4) {for (i[5] = 1; i[5] <= 9; i[5]++){int p5=0;for (int j5 = 0; j5 < 5; j5++) if (i[5] == i[j5]) p5 = 5;if (p5 != 5) {for (i[6] = 1; i[6] <= 9; i[6]++){int p6=0;for (int j6 = 0; j6 < 6; j6++) if (i[6] == i[j6]) p6 = 6;if (p6 != 6) {for (i[7] = 1; i[7] <= 9; i[7]++){int p7=0;for (int j7 = 0; j7 < 7; j7++) if (i[7] == i[j7]) p7 = 7;if (p7 != 7) {for (i[8] = 1; i[8] <= 9; i[8]++){int p8=0;for (int j8 = 0; j8 < 8; j8++) if (i[8] == i[j8]) p8 = 8;if (p8 != 8) {//printf("%d %d %d %d %d %d %d %d %d\n", i[0], i[1], i[2], i[3], i[4], i[5], i[6], i[7], i[8]);int a = 100 * i[0] + 10 * i[1] + i[2];int b = 100 * i[3] + 10 * i[4] + i[5];int c = 100 * i[6] + 10 * i[7] + i[8];double d1 = double(a) / b;double d2 = double(c) / b;if (d1 == 0.5 && d2 == 1.5){printf("%d %d %d\n", a, b, c);//system("pause");
                                                                        }}}}}}}}}}}}}}}}}}system("pause");return 0;
}

假设这三个数是 x : y : z  = 1 : 2 : 3，那么z最大可以为987，那么x最大到329（其实比329还要小几个数，因为数字重复了，懒得去抠了），那就好办了，x 从123循环到329，判重就完事了。

#include <iostream>
#include <string.h>
using namespace std;
int a[10];
//判重，且不能带0
int IsRepeat(int x) {memset(a, 0, sizeof(int) * 10);while (x) {int t = ++a[x % 10];if (t == 2 || a[0] == 1) return 1;x /= 10;}return 0;
}
int IsRepeat2(int x,int y,int z) {memset(a, 0, sizeof(int) * 10);while (x) {a[x % 10]++; a[y % 10]++; a[z % 10]++;if (a[0] > 0) return 1;x /= 10; y /= 10; z /= 10;}for (int i = 1; i < 10; i++) if (a[i] > 1) return 1;return 0;
}
int main() {//x,y,z 1:2:3,z最大可以为987，那么1最大到329for (int x = 123; x <= 329; x++) {if (IsRepeat(x)) {continue;}//如果x不重复，y,z 也不会重复int y = x * 2; int z = x * 3;if (!IsRepeat2(x, y, z)) {cout << x << " " << y << " " << z << endl;}}return 0;
}

 2.升级版

就是把原来的比例从1:2:3，变成了A:B:C。

思路还是差不多，中间continue判断，写的好像有点乱，但是确实不能少。

#include <iostream>
using namespace std;
bool IsRepeat(int x) {int a[10] = { 0 };while (x) {int t = a[x % 10]++;if (t == 2 || a[0] == 1)return true;x /= 10;}return false;
}
bool IsRepeat3(int x, int y, int z) {int a[10] = { 0 };while (x) {a[x % 10]++; a[y % 10]++; a[z % 10]++;if (a[0] > 0) return true;x /= 10; y /= 10; z /= 10;}for (int i = 1; i < 10; i++)if (a[i] > 1) return true;return false;
}
int main() {int A, B, C, flag = 0;cin >> A >> B >> C;for (int i = 123; i < 987; i++) {int j, k;if (IsRepeat(i))continue;if (i * B % A == 0) j = i * B / A;else continue;if (j > 987)continue;if (j*C%B == 0) {k = j * C / B;if (k > 987 || IsRepeat3(i, j, k))continue;flag = 1;cout << i << " " << j << " " << k << endl;}}if (!flag)cout << "No!!!" << endl;return 0;
}