Bound Found

题目描述：

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.

样例输入：

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

样例输出：

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

额，怎么说，这一题样例输入的输出结果应该有多种方法，评测应该会考虑到这一点，我们就不多说了

题目大意：

给定 n 个数，与 k 个 s 值，求区间总和的绝对值最接近 s 的一个区间，并输出它的总和绝对值、左端点、右端点。
1≤n≤100000

标准的尺取法，创造尺取法需要的单调性
代码如下：

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxn=100000+10;
pair<long long,int> p[maxn];
long long a[maxn];
long long juedui(long long x)
{
	if(x<0)return -x;
	return x;
}
int main()
{
	int n,m;
	while(1)
	{
		scanf("%d%d",&n,&m);
		if(m==0&&n==0)
		{
			break;
		}
		p[0]=make_pair(0,0);
		int i,j;
		long long x;
		long long sum=0;
		for(i=1;i<=n;i++)
		{
			scanf("%lld",&x);
			sum+=x;
			p[i]=make_pair(sum,i);
		}
		sort(p,p+n+1);
		long long t;
		while(m--)
		{
			scanf("%lld",&t);
			long long minn=99999999999,much;
			int start=0,over=1,l,r;
			while(over<=n&&minn)
			{
				long long num=p[over].first-p[start].first;
				if(juedui(num-t)<minn)
				{
					minn=juedui(num-t);
					l=p[start].second;
					r=p[over].second;
					much=juedui(num);
				}
				if(num<t)over++;
				else if(num>t)start++;
				if(start==over)over++; 
			}
			if(r<l)swap(r,l);
			printf("%lld %d %d\n",much,l+1,r);
		}
	}
	return 0;
}