Bound Found
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 5408 Accepted: 1735 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
解题心得：
- 给你一些列数，你选择一段连续的区间，区间总和的绝对值最接近给定的数，但是这些数可能是负数。
- 这个题主要的难点就是这些数可能是一个负数，有负数那么区间越长并不代表总和越大，这里就需要明确尺取法的两个特点
- 第一个特点就是尺取必须是选择连续的一段区间。
- 第二就是随着尺取长度的增加要有一个单调性，递增或者递减。
- 这个题明显不符合第二个特点随着区间长度的增加没有一个单调性，这个时候按照传统的尺取就没办法了。但是题目上并没有对于区间长度有什么要求，仅仅是要求总和绝对值和给定数相差最小，要求的是区间和，区间和可以从前缀和得到。这样就可以先得到前缀和，然后对于前缀和排序，这样就有了单调性，然后对于递增的前缀和用尺取法得最接近的给定数的区间。
#include <algorithm>
#include <stdio.h>
#include <climits>
using namespace std;
const int maxn = 1e5 + 100;
int t, n, m;
pair<int, int> p[maxn];
int main() {
while (scanf("%d%d", &n, &m) && n | m) {
p[0] = make_pair(0, 0);//要多插入一个为零的总和
for (int i = 1; i <= n; i++) {
int temp;
scanf("%d", &temp);
p[i] = make_pair(p[i - 1].first + temp, i);
}
sort(p, p + n + 1);
while (m--) {
scanf("%d", &t);
int l, r, sum = INT_MAX, ans_l, ans_r, ans = INT_MAX;
l = 0;r = 1;
while(r <= n) {
int temp = p[r].first - p[l].first;
if(abs(temp - t) < ans) {
ans = abs(temp-t);
sum = temp;
ans_l = min(p[l].second,p[r].second);
ans_r = max(p[l].second,p[r].second);
}
if(temp < t) r++ ;
else l++;
if(l == r)
r++;
}
printf("%d %d %d
", sum, ans_l + 1, ans_r);
}
}
return 0;
}