Bound Found POJ - 2566

 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

1.好题~~~虽然没做出来！
2.取数组的前缀和再排序再尺取。
3.关键在于转换

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstring>
using namespace std;

struct node{
    int sum,id;
    bool operator <(const node& i)const{
        return sum<i.sum;
    }
}a[100005];

int n,m;

void query(int t){
    int l=0,r=1,mid=2000000000;     //初始l和r，没有长度为0的区间！ 
    int ans,ansl,ansr;
    while(r<=n){
        int temp=a[r].sum-a[l].sum;
        if(abs(temp-t)<mid){
            mid=abs(temp-t);
            ans=temp;
            ansl=a[l].id;
            ansr=a[r].id;
        }
        if(temp>t) l++;
        else if(temp<t) r++;
        else break;
        if(l==r) r++;
    }
    if(ansl>ansr) swap(ansl,ansr);
    printf("%d %d %d\n",ans,ansl+1,ansr);
}

int main()
{    int x;
     while(~scanf("%d%d",&n,&m),n||m){
        a[0].id=0,a[0].sum=0;
        for(int i=1;i<=n;i++){
            scanf("%d",&x);
            a[i].sum=a[i-1].sum+x;
            a[i].id=i;
        }
        sort(a,a+n+1);   //一定要包括a[0]!
        while(m--){
            int q;scanf("%d",&q);
            query(q);
        }
    }
    return 0;
}