poj 2566 Bound Found (前缀和，尺取法（two pointer）)

Bound Found
Time Limit: 5000MS   Memory Limit: 65536K
Total Submissions: 2010   Accepted: 659   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　：给定一个长度为n的区间．然后给k次询问，每次一个数t,求一个区间[l,r]使得这个区间和的绝对值最接近t

没办法直接尺取.

先预处理出来前缀和

如果要找一对区间的和的绝对值最最近t

等价于找到两个数i和j,使得sum[i]-sum[j]的绝对值最接近t,且i<>j

那么对前缀和排序...然后尺取

因为答案要输出下标

所以之前先存一下下标．

然后对于i，j

所对应的区间为[min(pre[i].id,pre[j].id)+1,max(pre[i],id,pre[j].id)];

  1 /*************************************************************************
  2     > File Name: code/poj/2566.cpp
  3     > Author: 111qqz
  4     > Email: rkz2013@126.com 
  5     > Created Time: 2015年09月24日 星期四 22时10分08秒
  6  ************************************************************************/
  7 
  8 #include<iostream>
  9 #include<iomanip>
 10 #include<cstdio>
 11 #include<algorithm>
 12 #include<cmath>
 13 #include<cstring>
 14 #include<string>
 15 #include<map>
 16 #include<set>
 17 #include<queue>
 18 #include<vector>
 19 #include<stack>
 20 #include<cctype>
 21 #define y1 hust111qqz
 22 #define yn hez111qqz
 23 #define j1 cute111qqz
 24 #define ms(a,x) memset(a,x,sizeof(a))
 25 #define lr dying111qqz
 26 using namespace std;
 27 #define For(i, n) for (int i=0;i<int(n);++i)  
 28 typedef long long LL;
 29 typedef double DB;
 30 const int N=1E5+7;
 31 const int inf = 0x3f3f3f3f;
 32 int a[N];
 33 int n ,k;
 34 struct Q
 35 {
 36     int id;
 37     int sum;
 38 }pre[N];
 39 
 40 bool cmp(Q a,Q b)
 41 {
 42     if (a.sum<b.sum) return true;
 43     if (a.sum==b.sum&&a.id<b.id) return true;
 44     return false;
 45 }
 46 
 47 void solve ( int t)
 48 {
 49     int ansl,ansr,ans;
 50     int l = 0 ;
 51     int r = 1;
 52     int mi = inf;
 53     while ( r<= n )
 54     {
 55     int tmp = pre[r].sum - pre[l].sum;
 56     
 57     if (abs(tmp-t)<mi)
 58     {
 59         mi = abs(tmp-t);
 60         ansl = min(pre[l].id,pre[r].id)+1;
 61         ansr = max(pre[l].id,pre[r].id);
 62         ans = tmp;
 63     }
 64     if (tmp<t)
 65     {
 66         r++;
 67     }
 68     else
 69     if (tmp>t)
 70     {
 71         l++;
 72     }
 73     else break;
 74 
 75     if (l==r) r++;
 76     }
 77     printf("%d %d %d\n",ans,ansl,ansr);
 78 } 
 79 
 80 int main()
 81 {
 82   #ifndef  ONLINE_JUDGE 
 83    freopen("in.txt","r",stdin);
 84   #endif
 85     while (scanf("%d %d",&n,&k)!=EOF)
 86     {
 87     if (n==0&&k==0) break;
 88     pre[0].id = 0 ;
 89     pre[0].sum = 0 ;
 90     for ( int i = 1 ; i <= n ; i++)
 91     {
 92         scanf("%d",&a[i]);
 93         pre[i].sum = pre[i-1].sum + a[i];
 94         pre[i].id = i ;
 95        // cout<<pre[i].sum<<" ";
 96     }
 97 //    cout<<endl;
 98 
 99     sort(pre,pre+n+1,cmp);
100 //    for ( int i = 0 ; i <= n ; i++) cout<<pre[i].sum<<" "; cout<<endl;
101     for ( int i = 1 ; i <= k ; i++)
102     {
103         int x;
104         scanf("%d",&x);
105         solve(x);
106     }
107     }
108    
109  #ifndef ONLINE_JUDGE  
110   fclose(stdin);
111   #endif
112     return 0;
113 }
View Code

 

posted @ 2015-09-25 00:26  111qqz  阅读(190)  评论(0)  编辑  收藏  举报