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1011: [HNOI2008]遥远的行星

Time Limit: 10 Sec  Memory Limit: 162 MBSec  Special Judge
Submit: 4075  Solved: 1514
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Description

　　直线上N颗行星，X=i处有行星i,行星J受到行星I的作用力，当且仅当i<=AJ.此时J受到作用力的大小为 Fi->j=
Mi*Mj/(j-i) 其中A为很小的常量，故直观上说每颗行星都只受到距离遥远的行星的作用。请计算每颗行星的受力
，只要结果的相对误差不超过5%即可.

Input

　　第一行两个整数N和A. 1<=N<=10^5.0.01< a < =0.35,接下来N行输入N个行星的质量Mi,保证0<=Mi<=10^7

Output

　　N行，依次输出各行星的受力情况

Sample Input

5 0.3
3
5
6
2
4

Sample Output

0.000000
0.000000
0.000000
1.968750
2.976000

题解：乱搞就是了，刚开始5e5*1000以为不超时，找错边界了，后来AC了；

代码：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <list>
#include <set>
#include <map>
#define rep(i,a,b) for(int i = a;i <= b;++ i)
#define per(i,a,b) for(int i = a;i >= b;-- i)
#define mem(a,b) memset((a),(b),sizeof((a)))
#define FIN freopen("in.txt","r",stdin)
#define FOUT freopen("out.txt","w",stdout)
#define IO ios_base::sync_with_stdio(0),cin.tie(0)
#define mid ((l+r)>>1)
#define ls (id<<1)
#define rs ((id<<1)|1)
#define N 100005
#define INF 0x3f3f3f3f
#define INFF ((1LL<<62)-1)
using namespace std;
typedef long long LL;
typedef pair<int, int> PIR;
const double eps = 1e-8;
  
int n;
double A, m[N], ans[N], sum[N];
int main()
{
    //FIN;
    scanf("%d %lf", &n, &A);
    for(int i = 1; i <= n; ++i){
        scanf("%lf", &m[i]);
        sum[i] = sum[i-1] + m[i];
        //a[i] = A*double(i)+eps;
    }
    for(int i = 1;i <= n; ++i){
        int t = A*double(i)+eps;
        if(t > 200){
            int len = i-t/2;
            ans[i] += sum[t-200]*m[i]/len;
            for(int j = t-200;j <= t;++ j)
                ans[i] += m[i]*m[j]/(i-j);
        }
        else{
            for(int j = 1;j <= t;++ j)
                ans[i] += m[i]*m[j]/(i-j);
        }
    }
    for(int i = 1; i <= n; ++ i)    
        printf("%.6lf
", ans[i]);
    return 0;
}