Bound Found(尺取，poj2566， pair)

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本文介绍了一个编程挑战，目标是在给定的整数序列中找到一个子区间，使其和的绝对值与给定的目标值之差最小。通过排序和二分查找的方法解决了这一问题，并提供了完整的C++代码实现。

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Bound Found
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 5079 Accepted: 1618 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

大意：

求一个数组的子区间，使得其和的差值与t最小，任意一个。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;
int n, k;
typedef pair <long long int , int>p;
p sum[123123];
long long int a[123123];
long long int mabs(long long int x)
{
    if(x>=0)
        return x;
    else
        return -x;
}
int main()
{
    while(~scanf("%d %d", &n, &k)&&n+k)
    {
        sum[0] = p(0, 0);
        for(int i=1;i<=n;i++)
            {
                scanf("%lld", a+i);
                sum[i] = p(sum[i-1].first+a[i], i);
            }
            sort(sum, sum+1+n);//先按first，再按second
            while(k--)
            {
            int en = 1, st = 0;
            long long int ans = INF, t = 0, temp = 0, u=0, l=0, ka=0;
            scanf("%lld", &t);
            while(en<=n)
            {
                temp = sum[en].first - sum[st].first;
                if(mabs(t-temp)<ans)
                {
                    ans = mabs(t-temp);
                    ka = temp;
                    u = sum[en].second;
                    l = sum[st].second;
                }
                if(temp>t) st++;
                else if(temp<t)en++;
                else
                    break;
                if(st==en)
                    en++;
            }
            if(l>u)
                {
                    long long int kab = l;
                    l = u;
                    u = kab;
                }
            printf("%lld %lld %lld\n", ka, l+1, u);
        }
    }
    return 0;
}