POJ 2566

Bound Found

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 1445		Accepted: 487		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

 

按前缀和排序，尺取法解决

 1 #include 
 2 #include 
 3 #include 
 4 #include 
 5 #include 
 6 
 7 using namespace std;
 8 
 9 #define maxn 100005
10 #define INF 2000000000
11 
12 typedef pair<int,int> pii;
13 
14 int n,k;
15 pii a[maxn];
16 
17 int Abs(int x) {
18         return x > 0 ? x : -x;
19 }
20 
21 
22 void solve(int x) {
23         int sum = 0,s = 0,pos = 1,v,ans = INF,l,r;
24         //printf("ans = %d\n",ans);
25         for(; s <= n && pos <= n;) {
26                 int tem = a[pos].first - a[s].first;
27                 //printf("tem = %d\n",tem);
28                 if( Abs(tem - x) < ans) {
29                         ans = Abs(tem - x);
30                         l = a[s].second;
31                         r = a[pos].second;
32                         v = tem;
33                 }
34 
35                 if(tem > x) {
36                         ++s;
37                 } else if(tem < x) {
38                         ++pos;
39                 } else {
40                         break;
41                 }
42                 if(s == pos) ++pos;
43 
44         }
45         if(l > r) swap(l,r);
46 
47         printf("%d %d %d\n",v,l + 1,r);
48 }
49 
50 
51 
52 int main() {
53        // freopen("sw.in","r",stdin);
54 
55         while(~scanf("%d%d",&n,&k) ) {
56                 if(!n && !k) break;
57                 int sum = 0;
58                 a[0] = pii(0,0);
59                 for(int i = 1; i <= n; ++i) {
60                         int ch;
61                         scanf("%d",&ch);
62                         sum += ch;
63                         a[i] = make_pair(sum,i);
64 
65                 }
66 
67                 sort(a,a + n + 1);
68 
69                 for(int i = 1; i <= k; ++i) {
70                         int t;
71                         scanf("%d",&t);
72                         solve(t);
73                 }
74 
75         }
76         return 0;
77 }

View Code

 

转载于:https://www.cnblogs.com/hyxsolitude/p/3631980.html