2058: Find the Section

2058: Find the Section

---

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	569	128	Standard

In this task you are asked to deal with a lot of floating point numbers. Find the section in a number table that the sum of the numbers in it is the maximum.

Input

The first line of each test case is an integer N (<=100000). And the following lines have N floating point numbers. The last case have the N equal to zero, which you should not process.

Output

For each test case print "Case #T:" in the first line, where the T starts from 1. And the second line the sentence "The section from N1(th) to N2(th) have the max sum S", which means the numbers in the table from N1(th) to N2(th) have the sum S and S is the maximum you can find. N1, N2 are integers, and S is rounded to the nearest 0.001.

Note: If two sections have the same max sum, find the longer one and the one appears first. The length of the section can be zero which have a sum of zero, and it is from 0(th) to 0(th).

Sample Input

5
1 3.1 -2 7.35 -4
10
-5 -4 -3 -2 -1 0 1 2 3 4
0

Sample Output

Case #1:
The section from 1(th) to 4(th) have the max sum 9.450
Case #2:
The section from 6(th) to 10(th) have the max sum 10.000

 

Problem Source: coldcpp

#include<stdio.h>
int main()
{
    int n;
    int i,j;
    int l=1;
    while(scanf("%d",&n)==1&&n)
    {

        double a,s=1<<31,max=1<<31;
        int ti=1,tj=1,temp=1;
        for(i=1;i<=n;i++)
        {
            scanf("%lf",&a);
            if(s>=0) s+=a;//包含0 用'='
            else s=a,temp=i;
            if(s>max) max=s,ti=temp,tj=i;
            else if(s==max&&(i-temp>tj-ti)) ti=temp,tj=i;
        }
        if(max<0) max=ti=tj=0;
        printf("Case #%d:/n",l++);
        printf("The section from %d(th) to %d(th) have the max sum %.3lf/n",ti,tj,max);
    }
    return 0;
}