poj 2566 Bound Found（尺取法 好题）

时间：2021-08-03 12:26:04

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

 
尺取法，注意 inf 初始化
 
 #include<iostream>

 #include<cstdio>

 #include<cstring>

 #include<algorithm>

 #include<set>

 #include<map>

 using namespace std;

 #define N 106006

 #define inf 1<<30

 pair<int,int> g[N];

 int main()

 {

     int n,k;

     while(scanf("%d%d",&n,&k)==)

     {

         if(n== && k==)

           break;

         int sum=;

         g[]=make_pair(,);

         for(int i=;i<=n;i++){

             int x;

             scanf("%d",&x);

             sum=sum+x;

             g[i]=make_pair(sum,i);

         }

         sort(g,g+n+);

         while(k--){

             int val;

             scanf("%d",&val);

             int minn=inf;

             int ans,ansl=,ansr=;

             int s=,t=;

             for(;;){

                 if(t>n)

                   break;

                    if(minn==)

                      break;

                 int num=g[t].first-g[s].first;

                 if(abs(num-val)<minn){

                     minn=abs(num-val);

                     ans=num;

                     ansl=g[s].second;

                     ansr=g[t].second;

                 }

                  if(num<val)

                    t++;

                 if(num>val)

                   s++;

                 if(s==t)

                   t++;

             }

             if(ansl>ansr)

                 swap(ansl,ansr);

             printf("%d %d %d\n",ans,ansl+,ansr);

         }

     }

     return ;

 }

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