Bound Found
| Time Limit: 5000MS | Memory Limit: 65536K | |||
| Total Submissions: 5207 | Accepted: 1667 | 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.
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：给n和k，输入n个数和k个t，求一个连续子序列使得这个连续子序列的和最接近t，输出这个子序列的和和它的左右端点
解题思路：尺取法.要使用尺取法必须要保证数列的单调性，而这道题输入的n个数并不是单调的，然后借鉴了大佬的博客...这里用到了前缀和，预处理出前i个数的前缀和，和编号i一起放入pair中，然后根据前缀和从小到大排序，此时sum[r]-sum[l]就有了单调性。可以通过比较sum[r]-sum[l]与t的关系不断进行更新，如果sum[r]-sum[l]<t，说明和可以更大，所以r++；如果sum[r]-sum[l]>t，说明和可以更小，所以l++；如果sum[r]-sum[l]=t，必定是最小答案。由于序列不能为空，所以当l=r时，r++。更新答案的时候左右区间端点为乱序，输出的时候交换一下即可
因为前缀和不单调，所以需要先排序。在原数组开头添加0，求出前缀数组。题目即转化为在前缀数组中找pre[i]，pre[j]，两者之差最接近t，。对于每次找到的2个下标分别为i和j的2个数，所对应a的区间为[min(i, j) + 1, max(i, j)]。
那么前缀数组排序后，尺取法便可以求得最接近t的值。
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #define debu 6 using namespace std; 7 const int maxn = 1e5 + 50; 8 const int INF = 0x3f3f3f3f; 9 struct Node 10 { 11 int id, tot; 12 }; 13 Node a[maxn]; 14 int n, q, ans, ansl, ansr; 15 int cmp(Node a, Node b) 16 { 17 if (a.tot == b.tot) return a.id<b.id; 18 else return a.tot<b.tot; 19 } 20 int main() 21 { 22 while (scanf("%d%d", &n, &q) != EOF && (n + q)) 23 { 24 a[0].id = 0, a[0].tot = 0; 25 for (int i = 1; i <= n; i++) 26 { 27 int x; 28 scanf("%d", &x); 29 a[i].id = i; 30 a[i].tot = a[i - 1].tot + x; 31 } 32 sort(a, a + n + 1, cmp); 33 for (int i = 0; i<q; i++) 34 { 35 int t; 36 scanf("%d", &t); 37 int l = 0, r = 1, minx = INF; 38 while (l <= r && r <= n) 39 { 40 int tmp = a[r].tot - a[l].tot; 41 if (abs(tmp - t)<minx) 42 { 43 minx = abs(tmp - t); 44 ans = tmp; 45 ansl = a[l].id; 46 ansr = a[r].id; 47 } 48 if (tmp<t) r++; 49 else if (tmp>t) l++; 50 else break; 51 if (l == r) r++; 52 } 53 if (ansl>ansr) swap(ansl, ansr); 54 printf("%d %d %d ", ans, ansl + 1, ansr); 55 } 56 } 57 return 0; 58 }