POJ 2566 尺取法

 

POJ 2566 尺取法

分类: 技巧分析题   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
[cpp]  view plain copy
  1. #include<stdio.h>  
  2. #include<string.h>  
  3. #include<algorithm>  
  4. #include<limits.h>  
  5. #include<math.h>  
  6. using namespace std;  
  7. const int sz=110000;  
  8. int n;  
  9. pair<int,int>p[sz];  
  10. void find(int t)  
  11. {  
  12.     int i=0,j=1,ans=INT_MAX,l,v,r,temp,k;  
  13.     while(j<=n&&ans)  
  14.     {  
  15.         temp=p[j].first-p[i].first;  
  16.         if(abs(temp-t)<ans)  
  17.         {  
  18.             ans=abs(temp-t);  
  19.             v=temp;  
  20.             l=p[i].second;  
  21.             r=p[j].second;  
  22.         }  
  23.         if(temp<t) j++;  
  24.         if(temp>t) i++;  
  25.         if(i==j) j++;  
  26.     }  
  27.     if(l>r)  
  28.     {  
  29.         temp=l;  
  30.         l=r;  
  31.         r=temp;  
  32.   
  33.     }  
  34.     printf("%d %d %d\n",v,l+1,r);  
  35. }  
  36. int main()  
  37. {  
  38.     int q,i;  
  39.     while(scanf("%d %d",&n,&q)!=EOF)  
  40.     {  
  41.         if(!n&&!q) break;  
  42.         int sum=0,num;  
  43.         p[0]=make_pair(0,0);  
  44.         for(i=1;i<=n;i++)  
  45.         {  
  46.             scanf("%d",&num);  
  47.             sum+=num;  
  48.             p[i]=make_pair(sum,i);  
  49.         }  
  50.         sort(p,p+n+1);  
  51.         while(q--)  
  52.         {  
  53.             scanf("%d",&num);  
  54.             find(num);  
  55.         }  
  56.     }  
  57.     return 0;  
  58. }  

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