Bound Found POJ - 2566 (尺取)

最新推荐文章于 2021-08-30 23:38:14 发布

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本文介绍了一种用于解析可能源自外星的信号的算法。该算法接收一系列整数及目标值，通过查找连续子序列来确定其绝对和最接近目标值的范围。文章提供了输入输出示例，并附带了实现这一功能的C++代码。

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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

#include<stdio.h>
#include<string.h>
#include<map>
#include<vector>
#include<iostream>
#include<algorithm>
#include<queue>
#include<deque>
#include<math.h>
#define ll long long
const int inf=2e9;
using namespace std;
int a[100005];
struct node
{
    int x,sum;
}e[100005];
int k,n;
int cmp(node n1,node n2)
{
    return n1.sum<n2.sum;
}
void aa(int u)
{
    int l=0,r=1,l1,r1,ans=inf,kk;
    while(r<=n&&ans)
    {
        int dd=e[r].sum-e[l].sum;
        if(abs(dd-u)<=ans)
        {
            ans=abs(dd-u);
            kk=dd;
            l1=e[l].x,r1=e[r].x;
        }
        if(dd>u)
            l++;
        if(dd<u)
            r++;
        if(l==r) r++;
    }
    if(l1>r1) swap(l1,r1);
    printf("%d %d %d\n",kk,l1+1,r1);
}
int main()
{
    while(~scanf("%d%d",&n,&k)&&n+k)
    {
        int b;
        e[0].x=0;e[0].sum=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            e[i].x=i;
            e[i].sum=e[i-1].sum+a[i];
        }
        sort(e,e+n+1,cmp);
        while(k--)
        {
            scanf("%d",&b);
            aa(b);
        }
    }
}