POJ 2566 Bound Found（尺取+思维）

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

看别人说尺取适用于有单调性的问题，这个不可以直接用，想了半天，可能单调性说的是对于一个序列的和，如果左端点不动，右端点向右移动，那么和就一定会增加的，符合单调递增的特性。而这里可能减少，所以这道题不符合单调性。
如果要创造递增序列，可以把前缀和进行排序，这样左端点不动的情况下，右端点向右的过程中区间和一定会增加了。注意一下前缀和end和start不重合。

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;

ll func(ll x)
{
  return x>0?x:-x;
}
pair<int, int> q[100010];
int main()
{
  int n, k;
  while (scanf("%d%d", &n, &k)!=EOF){
    if (n==0 && k==0) break;
    q[0]=make_pair(0, 0);
    for (int i=1; i<=n; i++){
      int x;
      scanf("%d", &x);
      q[i]=make_pair(q[i-1].first+x, i);
    }
    sort(q, q+n+1);
    while (k--){
      ll t;
      scanf("%lld", &t);
      ll sum=0, ans;
      int minn=0x3f3f3f3f;
      int start=0, end=1;
      int l, r;
      while (end<=n){
        ll tt=q[end].first-q[start].first;
        if (func(tt-t)<minn){
          minn=func(tt-t);
          ans=tt;
          l=q[start].second;
          r=q[end].second;
        }
        if (tt>t) start++;
        else if (tt<t)  end++;
        else break;
        if (start==end) end++;
      }
      if (l>r)  swap(l, r);
      printf("%lld %d %d\n", ans, l+1, r);
    }
  }
  return 0;
}