Bound Found POJ - 2566（尺取法）

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
题意：找出一段连续的序列，使得这个序列的和的绝对值和t的差值最小。输出这段区间的绝对值和区间的左右端点。
思路：尺取的前提条件就是序列必须是单调的，否则前进和后退根本找不出规律来，就没有办法尺取了。但是如果堆这个数组排序的话，就不能保证这是连续的了。我们先把这段序列的前缀和数组求出来，然后将前缀和出现的序列号一同保存下来，按照由小到大排序。这样序列就是单调的了。然后两个指针开始尺取，如果当前区间求出的和小于t的话，说明还可以变大一点，那么就r++;如果当前区间求出的和大于t的话，说明还可以变小一点，那么就l++。注意这是前缀和，l和r不能相同。
代码如下：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;const int maxx=1e5+100;
struct node{int num;int pos;bool operator<(const node &a)const{return num<a.num;}
}p[maxx];
int a[maxx];
int n,m;int main()
{while(~scanf("%d%d",&n,&m)&&(n+m)){p[0].num=0;p[0].pos=0;//这里需要解释一下，因为这是前缀和，不是原来的数组，添加一个0是因为有可能从头开始取元素，那样的话，p[l].num需要为0才可以。for(int i=1;i<=n;i++) {scanf("%d",&a[i]);p[i].num=p[i-1].num+a[i];p[i].pos=i;}sort(p,p+1+n);//从0开始排序。while(m--){int t;scanf("%d",&t);int ans=inf,Ans;int L=0,R=0;int l=0,r=1;while(r<=n){int res=p[r].num-p[l].num;if(abs(res-t)<ans){ans=abs(res-t);Ans=res;L=p[l].pos;R=p[r].pos;}if(res<t) r++;else if(res>t) l++;else break;if(l==r) r++;}if(L>R) swap(L,R);printf("%d %d %d\n",Ans,L+1,R);}}return 0;
}

尺取就是需要多思考。
努力加油a啊，(^o)/~

Bound Found POJ - 2566（尺取法）相关推荐

Bound Found POJ - 2566（尺取法+前缀和创造区间变化趋势）
题意: 给定一个数组和一个值t,求一个子区间使得其和的绝对值与t的差值最小,如果存在多个,任意解都可行. 题目: Signals of most probably extra-terrestrial ...
Bound Found POJ - 2566 （尺取+前缀和）
题意就是给一串序列在给一个t 求一段区间的加和绝对值与t最小的值与区间左右端点是多少思路这道题一开始不知道如何去做用尺取法求区间但是因为区间中存在负数没有单调性没有特殊的特征若对区间求 ...
POJ 3320 尺取法，Hash，map标记
1.POJ 3320 2.链接:http://poj.org/problem?id=3320 3.总结:尺取法,Hash,map标记看书复习,p页书,一页有一个知识点,连续看求最少多少页看完所有知识 ...
尺取法 POJ 3601 Subsequence
题目传送门 1 /* 2 题意:求连续子序列的和不小于s的长度的最小值 3 尺取法:对数组保存一组下标(起点,终点),使用两端点得到答案 4 1. 记录前i项的总和,求[i, p)长度的最小值,用二分 ...
POJ 2566 Bound Found
题意 : 给你n个数字,这些数字可正可负,再给你个数字t, 求在这个数列中一个连续的子序列,和的绝对值与t相差最小: 数据范围较大, 考虑数字没有负数的情况, 能够想到用尺取法解决, (关于尺取法, ...
POJ 3061 (二分+前缀和or尺取法)
题目链接: http://poj.org/problem?id=3061 题目大意:找到最短的序列长度,使得序列元素和大于S. 解题思路: 两种思路. 一种是二分+前缀和.复杂度O(nlogn).有点 ...
尺取法 POJ 3320 Jessica's Reading Problem
题目传送门 1 /* 2 尺取法:先求出不同知识点的总个数tot,然后以获得知识点的个数作为界限, 更新最小值 3 */ 4 #include <cstdio> 5 #include &l ...
poj 3061（尺取法）
从长为n的全为正数的数组a中,找到个数最少的连续子数列使其和>=m example: n=10,m=15,a[10]={5 1 3 5 10 7 4 9 2 8},结果为2 解题思路:昨天做了一 ...
解题报告 (十三) 尺取法
文章目录尺取法解题报告 PKU 2100 Graveyard Design PKU 3061 Subsequence PKU 2739 Sum of Consecutive Prime Numbe ...

Bound Found POJ - 2566（尺取法）

Bound Found POJ - 2566（尺取法）相关推荐

最新文章

热门文章