POJ 2566 尺取法（进阶题）

Bound Found

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 4297		Accepted: 1351		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

题意：就是找一个连续的子区间，使它的和的绝对值最接近target, 区间内的元素可正可负。

思路：待会再补。。。

代码：

 #include "stdio.h"
 #include "stdlib.h"
 #include "iostream"
 #include "algorithm"
 #include "string"
 #include "cstring"
 #include "queue"
 #include "cmath"
 #include "vector"
 #include "map"
 #include "set"
 #define db double
 #define inf 0x3f3f3f
 #define mj
 //#define ll long long
 #define unsigned long long ull;
 using namespace std;
 const int mod = ;
 const int N=1e6+;
 const double eps = 1e-;
 typedef pair<int, int > pii;
 pii p[N];
 int n, m, k;
 void f(int k)
 {
     int l = , r = , ll, rr, v, mi = inf;
     while (l<=n&&r<=n&&mi!=)
     {
         int tmp=p[r].first - p[l].first;
         if (abs(tmp - k) < mi)
         {
             mi = abs(tmp - k);
             rr = p[r].second;
             ll = p[l].second;
             v = tmp;
         }
         if (tmp> k)
             l++;
         else if (tmp < k)
             r++;
         else
             break;
         if (r == l)
             r++;
     }
     if(ll>rr) swap(ll,rr);//因为ll和rr大小没有必然关系（）取绝对值，所以//要交换
     printf("%d %d %d\n", v, ll+, rr);
 }
 int main()
 {
     while (scanf("%d %d", &n, &m)==,n||m)
     {
         p[]=make_pair(, );
         for (int i = ; i <= n; i++)
         {
             scanf("%d", &p[i].first);
             p[i].first += p[i - ].first;
             p[i].second = i;
         }
         sort(p, p + n + );
         while (m--)
         {
             scanf("%d", &k);
             f(k);
         }
     }
     return ;
 }