CodeForces Round 176Div1 A Lucky permutation
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题解:
举个例子比较好理解,比如1 2 3 4 5 ,6,7,8,1对应其序号+1,2对应8,于是7,8,分别对应1,7,剩下还有3,4,5,6注意到可以重复上述步骤,于是这是一个可行序列,于是可以得出结论,若n%4==0则按上述规则是可行的,若n%4==1,则可以证明是可行的,其余数不可以
#include <cstdio>#include <cstring>using namespace std;int main(){ int flag[100010]; memset(flag,0,sizeof(flag)); int n; scanf("%d",&n); if(n%4==3||n%4==2) { printf("-1\n"); return 0; } int cnt=n/4; for(int i=1,j=1;i<=cnt;i++,j+=2) { flag[j]=j+1; flag[j+1]=n-j+1; flag[n-j+1]=n+1-flag[j]; flag[n-j]=n+1-flag[j+1]; } if(n%4==1) flag[n/2+1]=n/2+1; for(int i=1;i<=n;i++) { if(i==n) printf("%d\n",flag[i]); else printf("%d ",flag[i]); } return 0;}
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