标签:mem -- 模拟 std i++ mat cto src main
分析:码农模拟题,首先我们先求出对于0来说满足条件的数,然后在从集合当中筛选出任意两个都满足条件的数即可
1 /* 2 PROB:hamming 3 ID:wanghan 4 LANG:C++ 5 */ 6 #include "iostream" 7 #include "cstdio" 8 #include "cstring" 9 #include "string" 10 #include "vector" 11 #include "cmath" 12 #include "algorithm" 13 using namespace std; 14 const int maxn=1000+10; 15 int N,B,D; 16 vector<int> p; 17 int vis[maxn]; 18 string transform(int x,int y,string s) 19 { 20 string res=""; 21 int sum=0; 22 for(int i=0;i<s.length();++i) 23 { 24 if(s[i]==‘-‘) continue; 25 if(s[i]>=‘0‘&&s[i]<=‘9‘) 26 { 27 sum=sum*x+s[i]-‘0‘; 28 } 29 else 30 { 31 sum=sum*x+s[i]-‘A‘+10; 32 } 33 } 34 while(sum) 35 { 36 char tmp=sum%y; 37 sum/=y; 38 if(tmp<=9) 39 { 40 tmp+=‘0‘; 41 } 42 else 43 { 44 tmp=tmp-10+‘A‘; 45 } 46 res=tmp+res; 47 } 48 if(res.length()==0) res="0"; 49 if(s[0]==‘-‘) res=‘-‘+res; 50 return res; 51 } 52 string change(int x){ 53 string res=""; 54 while(x){ 55 int num=x%10; 56 res+=num+‘0‘; 57 x/=10; 58 } 59 int i=0,j=res.length()-1; 60 while(i<j){ 61 swap(res[i],res[j]); 62 i++,j--; 63 } 64 return res; 65 } 66 bool solve(int a,int b){ 67 string s1=transform(10,2,change(a)),s2=transform(10,2,change(b)); 68 string res1="",res2=""; 69 int cha; 70 if(s1.length()<B){ 71 cha=B-s1.length(); 72 for(int i=0;i<cha;i++) 73 res1+=‘0‘; 74 } 75 res1+=s1; 76 if(s2.length()<B){ 77 cha=B-s2.length(); 78 for(int i=0;i<cha;i++) 79 res2+=‘0‘; 80 } 81 res2+=s2; 82 int cnt=0; 83 for(int i=0;i<B;i++){ 84 if(res1[i]!=res2[i]){ 85 cnt++; 86 } 87 } 88 if(cnt>=D) 89 return true; 90 return false; 91 } 92 struct Node{ 93 int x; 94 string res; 95 }; 96 bool cmp(Node a,Node b){ 97 return a.res<b.res; 98 } 99 int main() 100 { 101 freopen("hamming.in","r",stdin); 102 freopen("hamming.out","w",stdout); 103 cin>>N>>B>>D; 104 p.push_back(0); 105 int pos=0; 106 int i=1; 107 int num=pow(2,B+1)-1; 108 while(i<num){ 109 if(solve(pos,i)){ 110 p.push_back(i); 111 } 112 i++; 113 } 114 memset(vis,0,sizeof(vis)); 115 int len=p.size(); 116 for(int i=0;i<len;i++){ 117 for(int j=i+1;j<len;j++){ 118 if(!vis[i]){ 119 if(vis[j]||!solve(p[i],p[j])){ 120 vis[j]=1; 121 } 122 } 123 } 124 } 125 vector<int> f; 126 for(int i=0;i<len;i++){ 127 if(!vis[i]){ 128 int zz=p[i]; 129 f.push_back(zz); 130 } 131 } 132 for(int i=0;i<N;i++){ 133 if(i%10==9){ 134 cout<<f[i]<<endl; 135 }else{ 136 if(i==N-1) 137 cout<<f[i]<<endl; 138 else 139 cout<<f[i]<<" "; 140 } 141 } 142 return 0; 143 }
标签:mem -- 模拟 std i++ mat cto src main
原文地址:http://www.cnblogs.com/wolf940509/p/7148858.html