poj 2566 Bound Found（取尺法，思路）

Bound Found

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 2719		Accepted: 826		Special Judge

Description

Signals of most probably extra-terrestrial origin have been received and digitalized by The Aeronautic and Space Administration (that must be going through a defiant phase: "But I want to use feet, not meters!"). Each signal seems to come in two parts: a sequence of n integer values and a non-negative integer t. We'll not go into details, but researchers found out that a signal encodes two integer values. These can be found as the lower and upper bound of a subrange of the sequence whose absolute value of its sum is closest to t. 

You are given the sequence of n integers and the non-negative target t. You are to find a non-empty range of the sequence (i.e. a continuous subsequence) and output its lower index l and its upper index u. The absolute value of the sum of the values of the sequence from the l-th to the u-th element (inclusive) must be at least as close to t as the absolute value of the sum of any other non-empty range.

Input

The input file contains several test cases. Each test case starts with two numbers n and k. Input is terminated by n=k=0. Otherwise, 1<=n<=100000 and there follow n integers with absolute values <=10000 which constitute the sequence. Then follow k queries for this sequence. Each query is a target t with 0<=t<=1000000000.

Output

For each query output 3 numbers on a line: some closest absolute sum and the lower and upper indices of some range where this absolute sum is achieved. Possible indices start with 1 and go up to n.

Sample Input

5 1
-10 -5 0 5 10
3
10 2
-9 8 -7 6 -5 4 -3 2 -1 0
5 11
15 2
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
15 100
0 0

Sample Output

5 4 4
5 2 8
9 1 1
15 1 15
15 1 15

题意：题意：给个t，求一个子区间与t的差的绝对值最小。

思路：n达到了100000，此题就不能用O（n²）枚举区间来做了。考虑连续区间上的xx问题，我们可以用尺取法。可以尺取法要求区间具有单调性，这里有正有负，怎么办？

很简单，我们要求任意区间的和就需要求出前缀数组，然后我们对前缀数组排序就可以了。  排完序后你找的任意两个点都会对应原序列的一个区间（因为会取绝对值，后面减去前面也没关系） 然后用尺取法推进，找出最小的一个即可。

代码：

#include 
#include 
#include 
#include 
#include 
using namespace std;
#define N 100050
#define INF 2000000010
int a[N];
struct Node
{
    int v,id;
}sum[N];
bool cmp(Node a,Node b)
{
    return a.vt) l++;
                else if(wansr) swap(ansl,ansr);
            printf("%d %d %d\n",anss,ansl+1,ansr);
        }
    }
    return 0;
}