poj 2566 尺取法

Bound Found

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 1732 Accepted: 574 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 10310 2-9 8 -7 6 -5 4 -3 2 -1 05 1115 2-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -115 1000 0

Sample Output

5 4 45 2 89 1 115 1 1515 1 15

Source

Ulm Local 2001

题目是，对于给定t，求出某段连续子序列的和使得其与t的差值最小。输出该和和下标范围。也就是找出区间使得 ||Sn-Sm|-t|最小，Sn代表序列前n项和。

|Sn-Sm|应该尽可能的接近t，因为这里是取绝对值，因此n和m的大小关系可以任意。我们可以先求出前缀和，然后进行从小到大排序,对于以Si,我们要找到j>i,Sj-Si最接近t的值，用尺取法进行查找。取前后两个指针i,j。Sj-Si<t的话，因为从小到大排好序了所以令j++，使区间和不断接近t，并把出现过的最小差值记录。Sj-Si>=t或j==n时，令i++,因此此时j继续增加得到的值是单调递增的，因此必然比之前出现的差值大，所以以i为起点的可能范围已经遍历完了，取下一个起点。

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cmath>using namespace std;#define maxn 100005int n, k;int t;int a[maxn];struct node{    int val;    int id;    bool operator < (const node &a) const    {        return val < a.val;    }} pre[maxn];int main(){    while(scanf("%d%d", &n, &k)!=EOF){        if(n+k==0) break;        for(int i = 1; i <= n; i++)            scanf("%d", a+i);        pre[0].val = 0, pre[0].id = 0;        for(int i = 1; i <= n; i++)            pre[i].val = pre[i-1].val+a[i], pre[i].id = i;        sort(pre, pre+n+1);        for(int i = 0; i < k; i++){        scanf("%d", &t);        int ll, rr, val, l, r;        l = 0, r = 1, val =-(0x3f3f3f3f);        while(l<r && r <= n){            int sum = pre[r].val-pre[l].val;            if(abs(sum-t) < abs(val-t)){                val = sum;                ll = l;                rr = r;            }            if(sum>=t){                l++;                if(l == r)                    r++;            }            else if(r == n)                l++;            else                r++;        }        printf("%d %d %d\n", val, min(pre[ll].id, pre[rr].id)+1, max(pre[ll].id, pre[rr].id));      }    }}