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

 
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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