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Bound Found
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 |
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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: [email protected]
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 }