Bound Found POJ - 2566（尺取法）

1.首先贡献一波模板

Subsequence通道：

#include<iostream>
#include<stdio.h>
#include<cmath>
using namespace std;
const int maxn  = 1e5+5;
int a[maxn],n,s,t,res;
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&s);
        for(int i = 0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        int l,r,sum;
        sum = 0;l = 0;r = 0;res = n+1;
        for(;;)
        {
            while(r<n&&sum<s)
            {
                sum+=a[r++];
            }
            if(sum < s)
                break;
            res = min(res,r - l);
            sum-=a[l++];
//            cout<<"sum = "<<sum<<endl;
        }
        if(res>n)
            printf("0\n");
        else
            printf("%d\n",res);
    }
    return 0;
}

2.好，接下来就是重点了，题目讲解。

Bound Found

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 6550		Accepted: 2112		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

题目通道：Bound Found

题意寻找不小于给定数的最接近给定数的区间和

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn = 100100;
const int INF = 1<<30;
int n,k,a[maxn],d,ans;
struct nod
{
    int v;
    int id;
};
struct nod sum[maxn];
bool cmp(nod a,nod b)
{
    if(a.v == b.v)
        return a.id>b.id;
    return a.v<b.v;
}

int main()
{
    while(scanf("%d%d",&n,&k)!=EOF&&(n||k))
    {
        sum[0].v = 0;sum[0].id = 0;
        for(int i = 1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum[i].v=sum[i-1].v+a[i];//区间内的和
            sum[i].id = i;//记录id号
        }
        for(int i = 0;i<k;i++)
        {
            ans = INF;
            scanf("%d",&d);
            int rex = 0,rey = 1,x,y,an;
            while(rex<=n&&rey<=n)
            {

                int sumx = abs(sum[rey].v - sum[rex].v);
                int de = abs(d - sumx);
                if(de<ans)//记录最小的与其差值
                {
                    ans = de;
                    x = sum[rex].id;y = sum[rey].id;
                    an = sumx;
                }
                if(sumx>d)
                    rex++;
                else if(sumx<d)
                    rey++;
                else
                    break;
                if(rex == rey)rey++;
            }
            if(x>y)//保证最小的在前面
            {
                int t = x;
                x = y;
                y = t;
            }
            printf("%d %d %d\n",an,x+1,y);//前面的那个Id是区间前的一个数，未记录所以应该x+1
        }
    }
    return 0;
}