Poj 2566(two pointers)

Bound Found

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 1338		Accepted: 443		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

好题。题意就是找一个连续的子区间，使它的和的绝对值最接近target。这题的做法是先预处理出前缀和，然后对前缀和进行排序，再用尺取法贪心的去找最合适的区间，要注意的是尺取法时首尾指针一定不能相同，因为这时区间相减结果为0，但实际上区间为空，这是不存在的，可能会产生错误的结果。处理时，把（0,0）这个点也放进数组一起排序，比单独判断起点为1的区间更方便。还有ans初始化的值INF一定要大于t的最大值。最后说说这个题最重要的突破口，对前缀和排序。为什么这么做是对的呢？以为这题是取区间的和的绝对值，所以所以用sum[r]-sum[l] 和 sum[l]-sum[r]是没有区别的。这样排序后，把原来无序的前缀和变成有序的了，就便于枚举的处理，并且不影响最终结果。

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 100000 + 5;
const int INF = 2000000000;
typedef  pair<int, int> P;
typedef long long LL;

P p[maxn];

int Abs(int x){return x<0?-x:x;}
int n;

void query(int tar){
    int s = 0,e = 1,Min = INF;
    int ansl,ansr,ansx;
    while(s <= n && e <= n){
        int tem = p[e].first-p[s].first;
        if(Abs(tem-tar) < Min){
            Min = Abs(tem-tar);
            ansx = tem;
            ansl = p[s].second;
            ansr = p[e].second;
        }
        if(tem > tar) s++;
        else if(tem < tar) e++;
        else break;
        if(s == e) e++;
    }
    if(ansl > ansr) swap(ansl,ansr);
    printf("%d %d %d\n",ansx,ansl+1,ansr);
}

int main(){
    int m;
    while(scanf("%d%d",&n,&m)){
        if(n == 0 && m == 0) break;
        p[0] = P(0,0);
        int sum = 0;
        for(int i = 1;i <= n;i++){
            int tem;
            scanf("%d",&tem);
            sum += tem;
            p[i] = P(sum,i);
        }
        sort(p,p+n+1);
        while(m--){
            int x;
            scanf("%d",&x);
            query(x);
        }
    }
    return 0;
}