Sicily 1800. Sequence

 

1800. Sequence

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

Given a sequence S of n numbers and two length bounds L and U.

Your task is to write a program to find a segment S[i], S[i+1], …, S[j] with minimum sum over all segments of S with length at least L and at most U.

Input

The input consists of multiple datasets. The end of the input is indicated by a line containing a zero.

The format of each dataset is as follows.

n

L U

S1 S2 … Sn

Here, all data items are integers. n is the length of the sequence (N≤32767). L and Uare two length bounds where L≤U. S1, S2, …, Sn are the sequence of n numbers.

Output

For each dataset, output the minimum sum of segment, in a separate line.

Sample Input

9
2 8
-3 2 -2 5 -4 1 -2 3 1
0

Sample Output

-5

Hint

the segment is -4 1 -2.

 

建立一棵线段树存储前缀和的最大值，注意从 0 开始

枚举区间右端点 i，查询 max(0,i-u) 到 i 范围内前缀和的最大值，用 i 处的前缀和减去它就是所求的最小值

其实一开始还想歪用 O(nlogn) 的分治法求最小连续子段和，枚举区间左右端点……这样就成了 O(n^2) 了（

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define rep(i,l,r) for(int i = l; i <= r; i++)
#define clr(x,y) memset(x,y,sizeof(x))
#define travel(x) for(Edge *p = last[x]; p; p = p -> pre)
using namespace std;
const int maxn = 32800;
inline int read(){
    int ans = 0, f = 1; char c = getchar();
    for(; !isdigit(c); c = getchar()) if (c == '-') f = -1;
    for(; isdigit(c); c = getchar()) ans = ans * 10 + c - '0';
    return ans * f;
}
struct Node{
    int l, r, s;
}t[maxn<<2];
int n, l, u, s[maxn];
void build(int u,int v,int w){
    t[w].l = u; t[w].r = v;
    if (u == v){
        t[w].s = s[u]; return;
    }
    int mid = (u + v) >> 1;
    build(u,mid,w<<1); build(mid+1,v,w<<1|1);
    t[w].s = max(t[w<<1].s, t[w<<1|1].s);
}
int query(int u,int v,int w){
    if (t[w].l == u && t[w].r == v) return t[w].s;
    int mid = (t[w].l + t[w].r) >> 1;
    if (v <= mid) return query(u,v,w<<1);
    else if (u > mid) return query(u,v,w<<1|1);
    else return max(query(u,mid,w<<1), query(mid+1,v,w<<1|1));
}
void work(){
    l = read(); u = read();
    s[0] = 0; rep(i,1,n) s[i] = s[i-1] + read();
    build(0,n,1);
    int ans = 0x7fffffff;
    rep(i,l,n) ans = min(ans, s[i] - query(max(0,i-u),i-l,1));
    printf("%d\n",ans);
}
int main(){
    n = read();
    while (n){
        work(); n = read();
    } 
    return 0;
}