poj2566

Bound Found

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 1959 Accepted: 636 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 10310 2-9 8 -7 6 -5 4 -3 2 -1 05 1115 2-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -115 1000 0

Sample Output

5 4 45 2 89 1 115 1 1515 1 15

代码如下：
/*踏实！！努力！！*/#include<iostream>#include<stdio.h>#include<cmath>#include<cstring>#include<map>#include<queue>#include<stack>#include<algorithm>using namespace std;#define INF 0x3f3f3f3fint n;pair<int,int> P[100005];void work(int k){    int ansl,ansr,ans,l=0,r=1;    int value=INF;    while(l<=n&&r<=n){        int temp=P[r].first-P[l].first;        if(abs(temp-k)<value){            ansl=P[l].second;            ansr=P[r].second;            ans=temp;            value=abs(temp-k);        }        if(temp<k) r++;        else if(temp>k) l++;        else break;        if(l==r) r++;    }    if(ansl>ansr) swap(ansl,ansr);    printf("%d %d %d\n",ans,ansl+1,ansr);}int main(){    int m,k;    while(scanf("%d%d",&n,&m)!=EOF){        if(!n&&!m) break;        long long sum=0;        P[0]=make_pair(0,0);        for(int i=1;i<=n;i++){            int x;            scanf("%d",&x);            sum+=x;            P[i]=make_pair(sum,i);        }        sort(P,P+1+n);        while(m--){            scanf("%d",&k);            work(k);        }    }    return 0;}