poj 2566 Bound Found（尺取法 好题）

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

Ulm Local 2001

 

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 ;
 }