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
按前缀和排序，尺取法解决
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath> using namespace std; #define maxn 100005
#define INF 2000000000 typedef pair<int,int> pii; int n,k;
pii a[maxn]; int Abs(int x) {
return x > ? x : -x;
} void solve(int x) {
int sum = ,s = ,pos = ,v,ans = INF,l,r;
//printf("ans = %d\n",ans);
for(; s <= n && pos <= n;) {
int tem = a[pos].first - a[s].first;
//printf("tem = %d\n",tem);
if( Abs(tem - x) < ans) {
ans = Abs(tem - x);
l = a[s].second;
r = a[pos].second;
v = tem;
} if(tem > x) {
++s;
} else if(tem < x) {
++pos;
} else {
break;
}
if(s == pos) ++pos; }
if(l > r) swap(l,r); printf("%d %d %d\n",v,l + ,r);
} int main() {
// freopen("sw.in","r",stdin); while(~scanf("%d%d",&n,&k) ) {
if(!n && !k) break;
int sum = ;
a[] = pii(,);
for(int i = ; i <= n; ++i) {
int ch;
scanf("%d",&ch);
sum += ch;
a[i] = make_pair(sum,i); } sort(a,a + n + ); for(int i = ; i <= k; ++i) {
int t;
scanf("%d",&t);
solve(t);
} }
return ;
}