首页 > 代码库 > POJ 2566 尺取法（进阶题）

POJ 2566 尺取法（进阶题）

2024-11-04 19:50:39   204人阅读

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 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
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：
 1 #include "stdio.h" 2 #include "stdlib.h" 3 #include "iostream" 4 #include "algorithm" 5 #include "string" 6 #include "cstring" 7 #include "queue" 8 #include "cmath" 9 #include "vector"10 #include "map"11 #include "set"12 #define db double13 #define inf 0x3f3f3f14 #define mj15 //#define ll long long16 #define unsigned long long ull;17 using namespace std;18 const int mod = 1000000007;19 const int N=1e6+5;20 const double eps = 1e-10;21 typedef pair<int, int > pii;22 pii p[N];23 int n, m, k;24 void f(int k)25 {26     int l = 0, r = 1, ll, rr, v, mi = inf;27     while (l<=n&&r<=n&&mi!=0)28     {29         int tmp=p[r].first - p[l].first;30         if (abs(tmp - k) < mi)31         {32             mi = abs(tmp - k);33             rr = p[r].second;34             ll = p[l].second;35             v = tmp;36         }37         if (tmp> k)38             l++;39         else if (tmp < k)40             r++;41         else42             break;43         if (r == l)44             r++;45     }46     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47     printf("%d %d %d\n", v, ll+1, rr);48 }49 int main()50 {51     while (scanf("%d %d", &n, &m)==2,n||m)52     {53         p[0]=make_pair(0, 0);54         for (int i = 1; i <= n; i++)55         {56             scanf("%d", &p[i].first);57             p[i].first += p[i - 1].first;58             p[i].second = i;59         }60         sort(p, p + n + 1);61         while (m--)62         {63             scanf("%d", &k);64             f(k);65         }66     }67     return 0;68 }

 

POJ 2566 尺取法（进阶题）


联系
我们