POJ 2566 尺取法

 

POJ 2566 尺取法

分类： 技巧分析题 2013-08-24 17:08  302人阅读  评论(0)  收藏  举报

: 5000MS		Memory Limit: 65536K
Total Submissions: 1348		Accepted: 448		Special Judge

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

 

题意：

输入 n m  之后输入n个数 

之后m个询问  对于每个询问 输入一个t    输出  三个数 ans l r  表示从l 到 r的所有数的和的绝对值最接近t 且输出这个和ans

 

 

思路： 求sum值之后排序用尺取法

 #include<stdio.h>
 #include<string.h>
 #include<algorithm>
 #include<limits.h>
 #include<math.h>
 using namespace std;
 const int sz=110000;
 int n;
 pair<int,int>p[sz];
 void find(int t)
 {
     int i=0,j=1,ans=INT_MAX,l,v,r,temp,k;
     while(j<=n&&ans)
     {
         temp=p[j].first-p[i].first;
         if(abs(temp-t)<ans)
         {
             ans=abs(temp-t);
             v=temp;
             l=p[i].second;
             r=p[j].second;
         }
         if(temp<t) j++;
         if(temp>t) i++;
         if(i==j) j++;
     }
     if(l>r)
     {
         temp=l;
         l=r;
         r=temp;

     }
     printf("%d %d %d\n",v,l+1,r);
 }
 int main()
 {
     int q,i;
     while(scanf("%d %d",&n,&q)!=EOF)
     {
         if(!n&&!q) break;
         int sum=0,num;
         p[0]=make_pair(0,0);
         for(i=1;i<=n;i++)
         {
             scanf("%d",&num);
             sum+=num;
             p[i]=make_pair(sum,i);
         }
         sort(p,p+n+1);
         while(q--)
         {
             scanf("%d",&num);
             find(num);
         }
     }
     return 0;
 }