Bound Found
| Time Limit: 5000MS | Memory Limit: 65536K | |||
| Total Submissions: 4297 | Accepted: 1351 | 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
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, 区间内的元素可正可负。
思路：待会再补。。。
代码：
#include "stdio.h"
#include "stdlib.h"
#include "iostream"
#include "algorithm"
#include "string"
#include "cstring"
#include "queue"
#include "cmath"
#include "vector"
#include "map"
#include "set"
#define db double
#define inf 0x3f3f3f
#define mj
//#define ll long long
#define unsigned long long ull;
using namespace std;
const int mod = ;
const int N=1e6+;
const double eps = 1e-;
typedef pair<int, int > pii;
pii p[N];
int n, m, k;
void f(int k)
{
int l = , r = , ll, rr, v, mi = inf;
while (l<=n&&r<=n&&mi!=)
{
int tmp=p[r].first - p[l].first;
if (abs(tmp - k) < mi)
{
mi = abs(tmp - k);
rr = p[r].second;
ll = p[l].second;
v = tmp;
}
if (tmp> k)
l++;
else if (tmp < k)
r++;
else
break;
if (r == l)
r++;
}
if(ll>rr) swap(ll,rr);//因为ll和rr大小没有必然关系（）取绝对值，所以//要交换
printf("%d %d %d\n", v, ll+, rr);
}
int main()
{
while (scanf("%d %d", &n, &m)==,n||m)
{
p[]=make_pair(, );
for (int i = ; i <= n; i++)
{
scanf("%d", &p[i].first);
p[i].first += p[i - ].first;
p[i].second = i;
}
sort(p, p + n + );
while (m--)
{
scanf("%d", &k);
f(k);
}
}
return ;
}