分析：SB题，列一个式子搞定
1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "string" 5 using namespace std; 6 long long n,k; 7 int main() 8 { 9 cin>>n>>k; 10 long long num=n/2; 11 num/=(k+1); 12 cout<<num<<" "<<(num*k)<<" "<<(n-num*(k+1))<<endl; 13 return 0; 14 }
分析：模拟，通过给定条件求出要求序列，注意一下不合法条件的判定
1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "string" 5 #include "cmath" 6 #include "vector" 7 using namespace std; 8 const int maxn=110; 9 int n,m; 10 int l[maxn],a[maxn]; 11 int vis[maxn]; 12 int main() 13 { 14 cin>>n>>m; 15 for(int i=1;i<=m;i++) 16 cin>>l[i]; 17 int num; 18 int flag=0; 19 memset(a,0,sizeof(a)); 20 for(int i=1;i<=m-1;i++){ 21 if(l[i+1]>l[i]){ 22 num=l[i+1]-l[i]; 23 }else{ 24 num=l[i+1]+n-l[i]; 25 } 26 if(a[l[i]]){ 27 if(num!=a[l[i]]){ 28 flag=1; break; 29 } 30 } 31 else a[l[i]]=num; 32 } 33 if(flag){ 34 cout<<"-1"<<endl; 35 return 0; 36 } 37 /*for(int i=1;i<=n;i++) 38 cout<<a[i]<<" "; 39 cout<<endl;*/ 40 memset(vis,0,sizeof(vis)); 41 for(int i=1;i<=n;i++) 42 vis[a[i]]++; 43 //int flag=0; 44 for(int i=1;i<=n;i++) 45 if(vis[i]>1){ 46 flag=1; break; 47 } 48 if(flag){ 49 cout<<"-1"<<endl; 50 }else{ 51 vector<int>h; 52 for(int i=1;i<=n;i++) 53 if(!vis[i]) 54 h.push_back(i); 55 int k=0; 56 for(int i=1;i<=n;i++) 57 if(!a[i]) 58 a[i]=h[k++]; 59 for(int i=1;i<=n;i++){ 60 if(i==n) cout<<a[i]<<endl; 61 else cout<<a[i]<<" "; 62 } 63 } 64 return 0; 65 }
分析：动态统计每个数出现的次数，一旦出现该数在某个位置出现的次数比A小，则淘汰
1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "string" 5 using namespace std; 6 const int maxn=1e6+10; 7 int vis[maxn],d[maxn]; 8 int n,A; 9 int main() 10 { 11 cin>>n>>A; 12 vis[A]=1; 13 for(int i=1;i<=n;i++){ 14 int x; 15 scanf("%d",&x); 16 if(d[x]<d[A]) vis[x]=1; 17 d[x]++; 18 } 19 int pos=-1; 20 for(int i=1;i<=maxn-10;i++) 21 if(d[i]>=d[A]&&!vis[i]){ 22 pos=i; break; 23 } 24 cout<<pos<<endl; 25 }