POJ2566-Bound Found

Bound Found

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 4057 Accepted: 1250 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 10310 2-9 8 -7 6 -5 4 -3 2 -1 05 1115 2-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -115 1000 0

Sample Output

5 4 45 2 89 1 115 1 1515 1 15

Source

Ulm Local 2001

题意：给出有n个数的一个序列，有q次询问，每次询问哪一段区间的绝对值最接近t

解题思路：这题不能用一般的尺取法来做，因为只有数列保证单调性才能用尺取法。可以先求一下前缀和，然后把前缀和从小到大进行排序，由于abs(sum[i]-sum[j])=abs(sum[j]-sum[i])，可以忽视数列前缀和的前后关系。此时，sum[r]-sum[l]有单调性。因此可以先比较当前sum[r]-sum[l]与t的差，并更新答案。果当前sum[r]-sum[l]<t,说明和还可以更大，r++。同理，如果sum[r]-sum[l]>t，说明和还可以更小，l++。如果sum[r]-sum[l]=t，必定是最小答案。由于序列不能为空，即l!=r，如果l==r则r++

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <stack>#include <cmath>#include <map>#include <bitset>#include <set>#include <vector>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int n, q, k, ansl, ansr, mi,ans;int a[100009];struct node{int id, sum;}x[100009];bool cmp(node a, node b){return a.sum < b.sum;}int main(){while (~scanf("%d%d", &n, &q) && (n + q)){x[0].id = 0, x[0].sum = 0;for (int i = 1; i <= n; i++) scanf("%d", &a[i]), x[i].id = i, x[i].sum = x[i - 1].sum + a[i];sort(x, x + n+1, cmp);while (q--){scanf("%d", &k);mi = INF;int l = 0, r = 1;while (mi != 0 && l <= n&&r <= n){int temp = x[r].sum - x[l].sum;if (abs(temp - k) < mi){mi = abs(temp - k);ans = temp;ansr = max(x[r].id, x[l].id);ansl = min(x[r].id, x[l].id) + 1;}if (temp > k) l++;else if(temp<k) r++;else break;if (l == r) r++;}printf("%d %d %d\n", ans, ansl, ansr);}}return 0;}