首页 > 代码库 > poj2566尺取变形

poj2566尺取变形

2024-09-16 21:37:56   215人阅读

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
题意：找连续的数使得和的绝对值和给定数差最小
题解：想了很久没想到，****因为尺取法必须单调数列才行，而且是求连续和****，先求前缀和，给其排序，再进行尺取
因为s不能等于t，当s==t时，t要++；sum>k时，向后移动s++，反之t++；
技术分享 
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007

using namespace std;

const int N=100000+5,maxn=10000+5,inf=0x3f3f3f3f;

struct edge{
   int v,id;
}a[N];
int n;

bool comp(const edge &a,const edge &b)
{
    return a.v<b.v;
}
void solve(int x)
{
    int s=0,t=1,ss,tt,sum,ans,minn=inf;
    while(s<=n&&t<=n&&minn!=0){
        sum=a[t].v-a[s].v;
        if(abs(sum-x)<minn)
        {
            minn=abs(sum-x);
            ans=sum;
            ss=a[s].id;
            tt=a[t].id;
        }
        if(sum>x)s++;
        else if(sum<x)t++;
        else break;
        if(t==s)t++;
    }
    if(ss>tt)swap(ss,tt);
    cout<<ans<<" "<<ss+1<<" "<<tt<<endl;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int k,s;
    while(cin>>n>>k,n||k){
        a[0].v=a[0].id=0;
        for(int i=1;i<=n;i++)
        {
            int p;
            cin>>p;
            a[i].v=a[i-1].v+p;
            a[i].id=i;
        }
        sort(a,a+n+1,comp);
        while(k--){
            cin>>s;
            solve(s);
        }
    }
    return 0;
}
View Code

 

poj2566尺取变形


联系
我们