标签:onclick str code ogre codeforce ssi pac bsp one
题意:将一列数划分为两个等差数列。
思路:首先,我要吹爆鸽巢原理!!!真的很强大的东西!!!
加入能完成题设操作,则前三个数中,必有至少两个数在同一序列,枚举三种情况(a1 a2,a2 a3,a1 a3分别为等差数列的前两项)。
注:枚举情况时,如果操作失败,则将已成功生成的等差数列末项划分到另一个数列试试。(稍作思考即可)
代码如下:
#include<cstdio> #include<iostream> #include<cstring> using namespace std; int n,arr[30010],book[30010]; bool judge(){ int ps,gap; int t1=1,t2; while(t1<=n&&book[t1]) t1++; t2=t1+1; if(t1>n) return false; while(t2<=n&&book[t2]) t2++; if(t2>n) return true; gap=arr[t2]-arr[t1]; ps=t2; for(int i=t2+1;i<=n;i++) if(!book[i]){ if(arr[i]-arr[ps]==gap) ps=i; else return false; } return true; } int main(){ scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&arr[i]); if(n==2) printf("%d\n%d",arr[1],arr[2]); else { int x[3]={arr[2]-arr[1],arr[3]-arr[2],arr[3]-arr[1]}; int flag=0; for(int i=0;i<3;i++){ int st=4,tmp=arr[3]; memset(book,0,sizeof(book)); if(i==0){ book[2]=book[1]=1; st=3; tmp=arr[2]; } else if(i==1) book[2]=book[3]=1; else book[1]=book[3]=1; for(int ii=st;ii<=n;ii++) if(arr[ii]==tmp+x[i]){ book[ii]=1; tmp=arr[ii]; } if(judge()){ flag=1; break; } else { int p=n; while(!book[p]) p--; book[p]=0; if(judge()){ flag=1; break; } } } if(flag){ for(int i=1;i<=n;i++) if(book[i]) printf("%d ",arr[i]); printf("\n"); for(int i=1;i<=n;i++) if(!book[i]) printf("%d ",arr[i]); } else printf("No solution"); } return 0; }
Two progressions(CodeForces-125D)[鸽巢原理]
标签:onclick str code ogre codeforce ssi pac bsp one
原文地址:https://www.cnblogs.com/xxmlala-fff/p/11617524.html