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 = 1000000007;
const int N=1e6+5;
const double eps = 1e-10;
typedef pair<int, int > pii;
pii p[N];
int n, m, k;
void f(int k)
{
    int l = 0, r = 1, ll, rr, v, mi = inf;
    while (l<=n&&r<=n&&mi!=0)
    {
        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
", v, ll+1, rr);
}
int main()
{
    while (scanf("%d %d", &n, &m)==2,n||m)
    {
        p[0]=make_pair(0, 0);
        for (int i = 1; i <= n; i++)
        {
            scanf("%d", &p[i].first);
            p[i].first += p[i - 1].first;
            p[i].second = i;
        }
        sort(p, p + n + 1);
        while (m--)
        {
            scanf("%d", &k);
            f(k);
        }
    }
    return 0;
}