Bound Found [POJ2566] [尺取法]

题意

给出一个整数列，求一段子序列之和最接近所给出的t。输出该段子序列之和及左右端点。

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

分析

这道题可以看得出来是要用尺取法，尺取法的关键是要找到单调性。而序列时正时负，显然是不满足单调性的。

如果记录下前缀和，并排好序，这样就满足单调性了，就可以使用前缀和了

代码

 #include<set>
 #include<map>
 #include<queue>
 #include<stack>
 #include<cmath>
 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #define RG register int
 #define rep(i,a,b)    for(RG i=a;i<=b;++i)
 #define per(i,a,b)    for(RG i=a;i>=b;--i)
 #define ll long long
 #define inf (1<<30)
 #define maxn 100005
 using namespace std;
 int n,k;
 int num[maxn];
 struct P{
     int id,s;
     inline int operator < (const P &a)const{
         return s==a.s?id<a.id:s<a.s;
     }
 }p[maxn];
 inline int read()
 {
     int x=,f=;char c=getchar();
     while(c<''||c>''){if(c=='-')f=-;c=getchar();}
     while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
     return x*f;
 }

 void solve()
 {
     int aim=read();
     int l=,r=,mn=inf,al,ar,ans;
     while(l<=n&&r<=n&&mn)
     {
         int cal=p[r].s-p[l].s;
         if(abs(cal-aim)<mn)
         {
             mn=abs(cal-aim);
             al=p[r].id,ar=p[l].id,ans=cal;
         }
         if(cal>aim)            ++l;
         else if(cal<aim)    ++r;
         else    break;
         if(l==r)    ++r;
     }
     if(al>ar)    swap(al,ar);
     printf("%d %d %d\n",ans,al+,ar);
 }

 int main()
 {
     while()
     {
         n=read(),k=read();
         if(!n&&!k)    return ;
         rep(i,,n)    num[i]=read();
         p[]=(P){,};
         rep(i,,n)    p[i].s=p[i-].s+num[i],p[i].id=i;
         sort(p,p++n);
         rep(i,,k)    solve();
     }
     return ;
 }