A narrow street is lined with tall buildings. An x foot long ladder is rested at the base of the building on the right side of the street and leans on the building on the left side. A y foot long ladder is rested at the base of the building on the left side of the street and leans on the building on the right side. The point where the two ladders cross is exactly c feet from the ground. How wide is the street?

Input Specification

Each line of input contains three positive floating point numbers giving the values of x, y, and c.

Output Specification

For each line of input, output one line with a floating point number giving the width of the street in feet, with three decimal digits in the fraction.

Sample Input

30 40 10
12.619429 8.163332 3
10 10 3
10 10 1

Sample Output

26.033
7.000
8.000
9.798

Status	In/Out	TIME Limit	MEMORY Limit	Submit Times	Solved Users	JUDGE TYPE
	stdin/stdout	3s	10240K	717	196	Standard

5
8 0 6 4 -1
-1

18

理解题意很重要，题目的意思是说 在m个中选   连续的n个atom能量值 使其最大。
程序中sumtemp，sum。sumtemp确定的是其左边界，sum确定其右边界。
sumtemp确定左边前n个数之和为负的最大的n，且第n个数显然为负，然后从n+1开始选数。
sum确定了右边界使其最大。
求和最大都可以用这种思路！！！！！！！！！！
举例
1，2，-4，4，2，-2，1，5，-6，5，-3，8，10
请看sumt1=1,sum=1
sumt2=2;sum=1+2=3;
sum3=sum2-4=-1;则sumtemp=0;sum不变。
sum4=sumtemp+4;sum<sum4,sum=4;
sum5=sumtemp+2；sum<sum5,sum=6
总之sum只有在sum<sumtemp时才修改。sumtemp<0则清0.
一维dp.
#include"stdio.h"
int main()
{
 freopen("s.txt","r",stdin);
  freopen("key.txt","w",stdout);
 int n;
 int a;
 while(scanf("%ld",&n),n!=-1)
 { 
      long long  sum=-0x7fffffff,sumtemp=-0x7fffffff;
   for(long i=0;i<n;i++)
   { 
    scanf("%d",&a);
    if(sumtemp>0)
   sumtemp+=a;
    else
   sumtemp=a;
    if(sumtemp>sum)
   sum=sumtemp;

   }
   printf("%lld\n",sum);

 }
 return 0;
}

A supermarket has a set Prod of products on sale. It earns a profit px for each product x∈Prod sold by a deadline dx that is measured as an integral number of time units starting from the moment the sale begins. Each product takes precisely one unit of time for being sold. A selling schedule is an ordered subset of products Sell ≤ Prod such that the selling of each product x∈Sell, according to the ordering of Sell, completes before the deadline dx or just when dx expires. The profit of the selling schedule is Profit(Sell)=Σ_x∈Sellpx. An optimal selling schedule is a schedule with a maximum profit.
For example, consider the products Prod={a,b,c,d} with (pa,da)=(50,2), (pb,db)=(10,1), (pc,dc)=(20,2), and (pd,dd)=(30,1). The possible selling schedules are listed in table 1. For instance, the schedule Sell={d,a} shows that the selling of product d starts at time 0 and ends at time 1, while the selling of product a starts at time 1 and ends at time 2. Each of these products is sold by its deadline. Sell is the optimal schedule and its profit is 80.

Input

A set of products starts with an integer 0 <= n <= 10000, which is the number of products in the set, and continues with n pairs pi di of integers, 1 <= pi <= 10000 and 1 <= di <= 10000, that designate the profit and the selling deadline of the i-th product. White spaces can occur freely in input. Input data terminate with an end of file and are guaranteed correct.

Output

For each set of products, the program prints on the standard output the profit of an optimal selling schedule for the set. Each result is printed from the beginning of a separate line.

Sample Input

4  50 2  10 1   20 2   30 1
7  20 1   2 1   10 3  100 2   8 2
5 20  50 10

Sample Output

80
185

Hint

The sample input contains two product sets. The first set encodes the products from table 1. The second set is for 7 products. The profit of an optimal schedule for these products is 185.

#include<iostream>
#include<cstdlib>
using namespace std;
#define MAX 10001
#define min(a,b) ((a)<(b) ? (a) : (b))
int father[MAX];
int p[MAX];
int result[MAX];
struct job{
     int value;
     int T;
  }JOB[MAX];
bool operator <(job job1,job job2)
 {
  if(job1.value>job2.value)
  return true;
  else
  return false;
 } 
 int find(int x)  //·µ»ØµÚ£Ø½ÚµãËùÊô¼¯ºÏµÄ¸ù½áµã
  {
int px=x;
while(p[px]>=0)
   px=p[px];
int tmp;
while(p[x]>=0)//ӦΪ³õֵΪ¸º
{
   tmp=p[x];
   p[x]=px;
   x=tmp;
}
return px;
}

   void UNION(int x,int y)
{
x=find(x);
y=find(y);
if(x==y)
   return ;
int tmp=p[x]+p[y];
if(p[x]>p[y])
{
   p[y]=tmp;
   p[x]=y;
}
else
{
   p[x]=tmp;
   p[y]=x;
}
}
 
  int main()
  {
  freopen("s.txt","r",stdin);
  freopen("key.txt","w",stdout);
  int num,temp=0;
  while(cin>>num)
  {
  memset(result,0,num);
  int i,l,j,k=0;
  for( i=0;i<num;i++)
  {
   cin>>JOB[i].value>>JOB[i].T;
   father[i]=i;
   p[i]=-1;
  }
  sort(JOB,JOB+num);
  for(i=0;i<num;i++)
  {
   j=find(min(JOB[i].T,num-1));//
   if(father[j]!=0)
       {
     k++;
     result[k]=i;
        l=find(father[j]-1);
        UNION(l,j);
        father[j]=father[l];
    }
  }
  i=0;
  for(j=1;j<=k;j++)
  { 
   i+=JOB[result[j]].value;}
    cout<<i<<endl;
   }

  //system("PAUSE");
  return   0;
  }
对着课本写得，自己都看不怎么懂。