【POJ】【2601】Simple calculations

推公式/二分法

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　　好题！

　　题解：http://blog.csdn.net/zck921031/article/details/7690288

　　这题明显是一个方程组……可以推公式推出来……

　　然而这太繁琐了！发现a[i]是满足单调性的话，我们就可以二分a[1]，递推出a[n+1]，进行验证……

　　思维复杂度比推公式低到不知哪里去了，真是一种优秀的算法（然而我想不到，并没有什么*用……）

 Source Code
 Problem:         User: sdfzyhy
 Memory: 736K        Time: 16MS
 Language: G++        Result: Accepted

     Source Code

     //PKUSC 2013 B
     //POJ 2601
     #include<vector>
     #include<cstdio>
     #include<cstring>
     #include<cstdlib>
     #include<iostream>
     #include<algorithm>
     #define rep(i,n) for(int i=0;i<n;++i)
     #define F(i,j,n) for(int i=j;i<=n;++i)
     #define D(i,j,n) for(int i=j;i>=n;--i)
     using namespace std;
     typedef long long LL;
     inline int getint(){
         int r=,v=; char ch=getchar();
         for(;!isdigit(ch);ch=getchar()) if (ch=='-') r=-;
         for(; isdigit(ch);ch=getchar()) v=v*-''+ch;
         return r*v;
     }
     const int N=;
     const double eps=1e-;
     /*******************template********************/

     int n;
     double a[N],c[N],ed;
     inline double check(double x){
         a[]=x;
         F(i,,n+) a[i]=*(a[i-]+c[i-])-a[i-];
         return a[n+];
     }
     int main(){
     #ifndef ONLINE_JUDGE
         freopen("B.in","r",stdin);
         freopen("B.out","w",stdout);
     #endif
         scanf("%d%lf%lf",&n,&a[],&ed);
         F(i,,n) scanf("%lf",&c[i]);
         double l=-,r=,mid;
         while(r-l>eps){
             mid=(l+r)/;
             if (check(mid)>ed) r=mid;
             else l=mid;
         }
         printf("%.2f\n",l);
         return ;
     }

Simple calculations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6646		Accepted: 3322

Description

There is a sequence of n+2 elements a₀, a₁, ..., a_n+1 (n <= 3000, -1000 <= a_i <=1000). It is known that a_i = (a_i-1 + a_i+1)/2 - c_i for each i=1, 2, ..., n.
You are given a₀, a_n+1, c₁, ... , c_n. Write a program which calculates a₁.

Input

The first line of an input contains an integer n. The next two lines consist of numbers a₀ and a_n+1 each having two digits after decimal point, and the next n lines contain numbers c_i (also with two digits after decimal point), one number per line.

Output

The output file should contain a₁ in the same format as a₀ and a_n+1.

Sample Input

1
50.50
25.50
10.15

Sample Output

27.85

Source

Ural State University collegiate programming contest 2000

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