POJ2566 Bound Found 2017-05-25 20:05 32人阅读 评论(0) 收藏

Bound Found

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 4056		Accepted: 1249		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

Source

Ulm Local 2001

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题目的意思是给出n个数，求一段区间和的绝对值最接近m的和及区间

思路：因为不存在单调，没法直接尺取，先处理波前缀和，在按前缀和排序，在用尺取法搞，注意这里尺取时的l和r不是区间端点

注意：inf取时不能0x3f3f3f3f，因为不够大

#include <iostream>
#include <cstring>
#include <cstdio>
#include <map>
#include <queue>
#include <cmath>
#include <algorithm>
#include <set>
using namespace std;
#define LL long long
const int inf=0x7fffffff;
int n,m,k;
int a[100005];
struct node
{
    int id,val;
} pre[100005];
bool cmp(node a,node b)
{
    return a.val<b.val;
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0)
            break;
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        pre[0].val=0;
        pre[0].id=0;
        for(int i=1; i<=n; i++)
            pre[i].val=pre[i-1].val+a[i],pre[i].id=i;

        sort(pre,pre+n+1,cmp);

        for(int i=0; i<m; i++)
        {
            scanf("%d",&k);
            int ans=inf;
            int ansl,ansr;
            int l=0,r=1;
            int sum;
            while(l<=n&&r<=n)
            {
                sum=pre[r].val-pre[l].val;
               if(abs(sum-k)<abs(ans-k))
                    {
                        ans=sum;
                        ansl=min(pre[l].id,pre[r].id)+1;
                        ansr=max(pre[l].id,pre[r].id);
                    }
                    if(sum<k)
                        r++;
                    else if(sum>k)
                        l++;
                    else
                        break;
                    if(l==r)
                        r++;
            }
            printf("%d %d %d\n",ans,ansl,ansr);
        }

    }
    return 0;
}