"bzoj1011" [HNOI2008] a distant planet

1011: [HNOI2008] Distant planet time limit:10 Sec Memory limit:162 mbsec Special Judge
submit:3711 solved:1375
[Submit] [Status] [Discuss] Description

On a straight line of n planets, there are planets I in the X=i, and Planet J is subjected to the force of the planet I, when and only as I<=aj. The size of the Force J is fi->j=
mi*mj/(J-i) where a is a very small constant, it is intuitively said that each planet is only affected by distant planets. Please calculate the force of each planet
, as long as the relative error of the result is not more than 5%.

Input

The first line is two integers n and a. 1<=n<=10^5.0.01< A < = 0.35, next n lines enter the mass Mi of n planets, guaranteeing 0<=mi<=10^7

Output

n rows, sequentially outputting the forces of each planet

Sample Input5 0.3
3
5
6
2
4Sample Output0.000000
0.000000
0.000000
1.968750
2.976000HINT

The exact result should be 0 0 0 2 3, but the result error of the sample output is not more than 5%, also

"Spit Groove"

The provincial election actually exam such a hole problem ...

"As long as the relative error of the result is not more than 5%," This is the key word.

Because when the planets are very far apart, the force is very small, it can be estimated that the distance from the former 1-now planet to the current planet is NOW/2, O (1) solved. (Is this a greedy strategy?) ）

How to Improve IQ??? It's a permanent topic.

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <cmath>5#include <cstdlib>6#include <ctime>7#include <algorithm>8 using namespacestd;9 #defineMAXN 100010Ten #defineEPS 1e-8 One intN; A DoubleA,M[MAXN],SUM[MAXN],ANS[MAXN]; -InlineintRead () - { the     intx=0, f=1;CharCh=GetChar (); -      while(!isdigit (CH)) {if(ch=='-') f=-1; Ch=GetChar ();} -      while(IsDigit (CH)) {x=x*Ten+ch-'0'; Ch=GetChar ();} -     returnx*F; + } - intMain () + { AN=read (); scanf"%LF", &a); sum[0]=0.000000; at      for(intI=1; i<=n;i++) {scanf ("%LF", &m[i]); sum[i]=sum[i-1]+m[i];} -      for(intI=1; I<=min ( -, n); i++) -     { -         intR=eps+floor (A *i); -          for(intj=1; j<=r;j++) ans[i]+=m[i]*m[j]/(i-j); -     } in      for(intI=2001; i<=n;i++) -     { to         intR=eps+floor (A *i); +ans[i]+=sum[r]*m[i]/(Double) (I-(int) r/2); -     } the      for(intI=1; i<=n;i++) printf ("%6lf\n", Ans[i]); *     return 0; $}

"bzoj1011" [HNOI2008] a distant planet