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.
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
按前缀和排序，尺取法解决
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 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 }