Bound Found
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
思路:
尺取法+前缀和
完整代码:
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#define INF 0x3f3f3f3f
using namespace std;
int n,k,m;
struct node
{
int sum,c;
}v[100005];
bool cmp(node a,node b)
{
return a.sum<b.sum;
}
void Swap(int &a,int &b)
{
int t=a;
a=b;
b=t;
return ;
}
void take(int x) //尺取法
{
int l=0,r=1,minn=INF;
int ll,rr,ans;
while(l<=n && r<=n && minn!=0)
{
int t=v[r].sum-v[l].sum;
if(abs(t-x)<minn)
{
ll=v[l].c;
rr=v[r].c;
ans=t;
minn=abs(t-x);
}
if(t>x)
{
l++;
}
else if(t<x)
{
r++;
}
else
{
break;
}
if(l==r)
{
r++;
}
}
if(ll>rr)
{
Swap(ll,rr);
}
cout<<ans<<" "<<ll+1<<" "<<rr<<endl;
}
int main()
{
while(cin>>n>>k)
{
if(n==0&&k==0)
{
break;
}
memset(v,0,sizeof(v));
for(int i=1;i<=n;i++)
{
cin>>m;
v[i].c=i; //前缀和
v[i].sum=v[i-1].sum+m;
}
sort(v,v+n+1,cmp);
while(k--)
{
int x;
cin>>x;
take(x);
}
}
return 0;
}