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 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

Source

Ulm Local 2001好題。題意就是找一個連續的子區間，使它的和的絕對值最接近target。這題的做法是先預先處理出首碼和，然後對首碼和進行排序，再用尺取法貪心的去找最合適的區間，要注意的是尺取法時首尾指標一定不能相同，因為這時區間相減結果為0，但實際上區間為空白，這是不存在的，可能會產生錯誤的結果。處理時，把（0,0）這個點也放進數組一起排序，比單獨判斷起點為1的區間更方便。還有ans初始化的值INF一定要大於t的最大值。最後說說這個題最重要的突破口，對首碼和排序。為什麼這麼做是對的呢？以為這題是取區間的和的絕對值，所以所以用sum[r]-sum[l]
和 sum[l]-sum[r]是沒有區別的。這樣排序後，把原來無序的首碼和變成有序的了，就便於枚舉的處理，並且不影響最終結果。

#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;}