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:给你n个数,让你找一个子区间使得该子区间之和的绝对值和t的差尽可能小,输出该区间的和,以及区间的左右端点。
题解:因为数组中的数有正有负,故不满足尺取单调性的要求,我们可以考虑求前缀和,因为两个前缀和相减对应的是任意一个区间的和,因为有绝对值,所以不用考虑端点的先后顺序,我们将所有前缀和从小到大排序,然后按照尺取遍历即可。
#include<math.h>
#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
int a[100005],l,r,n,k;
ll sm,ans,t;
ll abss(ll x)
{
if(x<0)
x=-x;
return x;
}
struct node
{
ll s,p;
}sum[100005];
bool comp(node a,node b)
{
return a.s<b.s;
}
int main(void)
{
int i;
while(scanf("%d%d",&n,&k)!=EOF)
{
if(n==0 && k==0)
break;
sum[0].p=sum[0].s=0;
for(i=1;i<=n;i++)
scanf("%d",&a[i]),sum[i].s=sum[i-1].s+a[i],sum[i].p=i;
sort(sum,sum+n+1,comp);
while(k--)
{
ans=1000000000000000;
scanf("%lld",&t);
l=0;r=1; ll ans1; int ls,rs;
while(l<=n && r<=n)
{
ll tmp=abss(sum[l].s-sum[r].s);
ll tmp1=abss(tmp-t);
if(tmp1<ans)
{
ans=tmp1;
ans1=tmp;
ls=sum[l].p;
rs=sum[r].p;
}
if(tmp>t) l++;
else if(tmp<t) r++;
else break;
if(l==r) r++;
}
if(ls>rs) swap(ls,rs);
printf("%lld %d %d\n",ans1,ls+1,rs);
}
}
return 0;
}