POJ2566 ZOJ1964 Bound Found【前缀和+排序+尺取法】

Bound Found

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 4085		Accepted: 1266		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

问题链接：POJ2566 ZOJ1964 Bound Found。

题意简述：给出一个整数序列，求其子序列之和最接近所给定的t，输出该子序列和及两端序号。

问题分析：

序列的各个元素有正有负，事情比较麻烦一些。

需要先计算前缀和。两个前缀和的差即为一个子序列的和。

把前缀和及其下标进行排序，就可以用尺取法（序列前缀和单调递增才可以使用）来求最接近的子序列和的值及两端的序号。

程序说明：（略）

AC的C++语言程序如下：

/* POJ2566 ZOJ1964 Bound Found */

#include <iostream>
#include <algorithm>
#include <limits.h>
#include <stdio.h>

using namespace std;

const int N = 100000;
pair<int, int> prefixsum[N+1];

void find(int t, int n)
{
    int delta = INT_MAX;
    int left, right, ans;
    int start=0, end=1, sum;

    while(end <= n && delta) {
        sum = prefixsum[end].first - prefixsum[start].first;
        if(abs(sum - t) < delta) {
            delta = abs(sum - t);

            ans = sum;
            left = prefixsum[start].second;
            right = prefixsum[end].second;
        }
        if(sum < t)
            end++;
        if(sum > t)
            start++;
        if(start == end)
            end++;
    }

    if(left > right) {
        sum = left;
        left = right;
        right = sum;
    }

    printf("%d %d %d
", ans, left+1, right);
}

int main()
{
    int n, k, a, sum, t;

    while(scanf("%d%d", &n, &k) != EOF && (n || k)) {
        prefixsum[0] = make_pair(0, 0);

        sum = 0;
        for(int i=1; i<=n; i++) {
            scanf("%d", &a);

            sum += a;
            prefixsum[i] = make_pair(sum, i);
        }

        sort(prefixsum, prefixsum + n + 1);

        while(k--) {
            scanf("%d", &t);
            find(t, n);
        }
    }

    return 0;
}