Bound Found
Time Limit: 5000MS | Memory Limit: 65536K | |||
Total Submissions: 4384 | Accepted: 1377 | 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.
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
/* * @Author: Lyucheng * @Date: 2017-08-02 10:22:54 * @Last Modified by: Lyucheng * @Last Modified time: 2017-08-02 15:57:35 */ /* 题意;给你一个序列,然后有k次查询,让你找一个子序列绝对值最接近t的序列 思路:尺取,如果想要使用尺取,要保证数列的单调性,但是序列中有负数,要给前缀排序 */ #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <time.h> #define MAXN 100005 #define INF 2147483640 using namespace std; struct Node{ int id; int val; bool operator < (const Node & other) const { return val<other.val; } }node[MAXN]; int n,k; int t; int a; int l,r; int res_l,res_r; int res; int _min; int main(){ // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); while(scanf("%d%d",&n,&k)!=EOF){ node[0].id=0; node[0].val=0; for(int i=1;i<=n;i++){ scanf("%d",&a); node[i].val=node[i-1].val+a; node[i].id=i; } sort(node,node+n+1); while(k--){ scanf("%d",&t); l=0,r=1; _min=INF; while(r<=n&&_min){ int pos=abs(node[r].val-node[l].val); if(abs(pos-t)<=_min){ _min=abs(pos-t); res=pos; res_l=node[l].id; res_r=node[r].id; } if (pos<t) r++; if (pos>t) l++; if (l==r) r++; } if(res_l>res_r) swap(res_l,res_r); printf("%d %d %d ",res,res_l+1,res_r); } } return 0; }