等边三角形
有如下一个题目:
我就直接for循环套,暴力求解。
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define test cout<<"***"<<endl 4 int n; 5 int an[15]; 6 int main(){ 7 cin>>n; 8 int cnt = 0; 9 memset(an,0,sizeof(an)); 10 int a,b,c,d,e,f,g,h,i; 11 for(int i1=1;i1<=9;i1++){ 12 // test; 13 a = i1; 14 an[i1] = 1; 15 for(int j=i1+1;j<=9;j++){ 16 f = j; 17 an[j] = 1; 18 for(int k=j+1;k<=9;k++){ 19 i = k; 20 an[k] = 1; 21 for(int p = 1;p<=9;p++){ 22 if(an[p]==0){ 23 b = p; 24 an[p] = 1; 25 for(int p1 = b;p1<=9;p1++){ 26 if(an[p1]==0){ 27 d = p1; 28 an[p1] = 1; 29 for(int p2=1;p2<=9;p2++){ 30 if(an[p2]==0){ 31 c = p2; 32 an[p2] = 1; 33 for(int p3 = c;p3<=9;p3++){ 34 if(an[p3]==0){ 35 e = p3; 36 an[p3] = 1; 37 for(int p4 =1;p4<=9;p4++){ 38 if(an[p4]==0){ 39 g = p4; 40 an[p4] = 1; 41 for(int p5 =g;p5<=9;p5++){ 42 if(an[p5]==0){ 43 h = p5; 44 an[p5]= 1; 45 int x=a+b+d+f; 46 int y=f+g+h+i; 47 int z=i+e+c+a; 48 if(x==y&&y==z&&x==n){ 49 cnt++; 50 // cout<<a<<" "<<b<<" "<<c<<" "<<d<<" "<<e<<" "<<f<<" "<<g<<" "<<h<<" "<<i<<endl; 51 } 52 an[p5]=0; 53 } 54 } 55 an[p4] = 0; 56 } 57 } 58 an[p3] = 0; 59 } 60 } 61 an[p2] = 0; 62 } 63 } 64 an[p1] = 0; 65 } 66 } 67 an[p] = 0; 68 } 69 } 70 an[k] = 0; 71 } 72 an[j] = 0; 73 } 74 an[i1]=0; 75 } 76 if(cnt) 77 cout<<cnt<<endl; 78 else 79 cout<<"No"<<endl; 80 return 0; 81 }
下面是后台测试数据(过了说明代码就行了):
23 2
18 No
19 4
20 6
21 4