Bound Found POJ - 2566 尺取

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本文介绍了一种算法，用于解决给定序列中找到一个子区间的问题，该子区间的和的绝对值最接近指定的目标值。通过预计算所有可能的子区间和并进行排序，可以有效地找到最佳子区间。

Bound Found

POJ - 2566

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

题意：求和的绝对值最接近t的区间

既然要找到一个子区间使得和最接近t的话，那么不断地找比当前区间的和更大的区间，如果区间和已经大于等于t了，那么不需要在去找更大的区间了，因为其和与t的差值更大，然后区间左端点向右移动推进即可。所以，首先根据计算出所有的区间和，

排序之后按照上面的思路求解即可。

#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
#define ll long long 
const int INF = 0x3f3f3f3f;
const int N = 1e5 + 7;
struct S {
	ll sum, id;
	bool operator< (const S& a) const {
		if (sum == a.sum) return id < a.id;
		return sum < a.sum;
	}
} sum[N];
ll a[N];
template<typename T>
T myabs(T x) {
	return x > 0 ? x : -x;
} 
int main() {
	int n, k;
	while (~scanf ("%d %d", &n, &k) && (n + k)) {
		sum[0].sum = 0, sum[0].id = 0;
		for (int i = 1; i <= n; ++i) {
			scanf ("%lld", &a[i]);
			sum[i].sum = sum[i-1].sum + a[i];
			sum[i].id = i;
		}
		sort(sum, sum+n+1);
		while (k--) {
			ll t;
			scanf ("%lld", &t);
			int tail = 0, head = 1, l = 0, r = 0;
			ll sum1 = 0, ans, tmp = INF;
			while (head <= n) {
				sum1 = sum[head].sum - sum[tail].sum;
				if (myabs(t - sum1) < tmp) {
					tmp = myabs(t - sum1);
					ans = sum1, r = sum[head].id, l = sum[tail].id;
				}
				if (sum1 < t) ++head;
				else if (sum1 > t) ++tail;
				else break;
				if (head == tail) ++head;
			}
			if (l > r) swap(l, r);
			printf ("%lld %d %d\n", ans, l+1, r);
		}
	}
	return 0;
}