poj 2566 Bound Found

Bound Found

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 2773		Accepted: 845		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

Source

Ulm Local 2001

尺取法的进化版本！！！

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 100000 + 5;
const int inf = 0x3f3f3f3f;
int A[maxn],N,K;
struct Sum
{
    int s,p;
    inline bool operator < (const Sum &b)   const   { return s < b.s; }
} S[maxn];
inline void Qry(int x)
{
    int l = 0,r = 1,mn = inf;
    int rc,rc_l,rc_r;
    while(l <= N && r <= N)
    {
        int t = S[r].s - S[l].s;
        if(abs(t - x) < mn)
        {
            rc_l = S[l].p,rc_r = S[r].p;
            rc = t;     mn = abs(t - x);
        }
        if(t > x)       ++ l;
        else if(t < x)  ++ r;
        else            break;
        if(l == r)      ++ r;
    }
    if(rc_l > rc_r) swap(rc_l,rc_r);
    printf("%d %d %d\n",rc,rc_l + 1,rc_r);
}
int main()
{
    while(scanf("%d%d",&N,&K), N || K)
    {
        S[0].s = 0; S[0].p = 0;
        for(int i = 1;i <= N;i ++)  scanf("%d",A + i),S[i].s = S[i - 1].s + A[i],S[i].p = i;
        sort(S,S + 1 + N);
        while(K --)
        {
            int T;  scanf("%d",&T);
            Qry(T);
        }
    }
    return 0;
}