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poj2566 Bound Found
Bound Found
| Time Limit: 5000MS | Memory Limit: 65536K | |||
| Total Submissions: 1651 | Accepted: 544 | 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 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
思路:sum[i][j]=sum[0][j]-sum[0][i-1],所以可以把部分和问题转换成求两个和之间的差最接近T的问题
但是差可能有负也有正,那就把和排序一遍,这样就只能得到非负数差,可以用尺取,记录下编号小的在前就行了
#include <cstdio>#include <algorithm>using namespace std;const int maxn=100005;int n,k,T;typedef pair<long long ,int> P;P sum[maxn];int nsts,nste;long long nstt;long long calc(int s,int e){ return sum[e].first-sum[s].first;}int main(){ while(scanf("%d%d",&n,&k)==2&&n&&k){ long long s=0; sum[0].first=0; sum[0].second=0;//这个不能在结果中出现,为了使得0存在而加入,是不含元素的和 for(int i=1;i<=n;i++){ int tmp; scanf("%d",&tmp); s+=tmp; sum[i].first=s; sum[i].second=i; } nsts=nste=1;nstt=sum[1].first; sort(sum,sum+n+1); for(int i=0;i<k;i++){ int l=0,r=1; scanf("%d",&T); while(l<r&&r<=n){ long long tmp=calc(l,r); if(abs(tmp-T)<abs(nstt-T)){ nstt=tmp; nsts=min(sum[l].second,sum[r].second)+1; nste=max(sum[l].second,sum[r].second); } if(tmp>T&&l<r-1){ l++; } else { r++; } } printf("%I64d %d %d\n",nstt,nsts,nste); } } return 0;}
poj2566 Bound Found
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