Codeforces Round #211 (Div. 2)B. Fence

B. Fence
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

There is a fence in front of Polycarpus's home. The fence consists of n planks of the same width which go one after another from left to right. The height of the i-th plank is hi meters, distinct planks can have distinct heights.

Fence for n = 7 and h = [1, 2, 6, 1, 1, 7, 1]

Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly k consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such kconsecutive planks that the sum of their heights is minimal possible.

Write the program that finds the indexes of k consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic).

Input

The first line of the input contains integers n and k (1 ≤ n ≤ 1.5·105, 1 ≤ k ≤ n) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers h1, h2, ..., hn (1 ≤ hi ≤ 100), where hi is the height of the i-th plank of the fence.

Output

Print such integer j that the sum of the heights of planks jj + 1, ..., j + k - 1 is the minimum possible. If there are multiple such j's, print any of them.

Examples
input
7 3
1 2 6 1 1 7 1
output
3
Note

In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8.

求连续的k个数和的最小值,要求输出下标

前缀和瞎搞  O(n)

/* ***********************************************
Author        :guanjun
Created Time  :2016/10/7 16:24:54
File Name     :cf211b.cpp
************************************************ */
#include <bits/stdc++.h>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
    int x,y;
};
struct cmp{
    bool operator()(Node a,Node b){
        if(a.x==b.x) return a.y> b.y;
        return a.x>b.x;
    }
};

bool cmp(int a,int b){
    return a>b;
}
int dp[200010];
int a[200010];
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int n,k,ans;
    while(cin>>n>>k){
        for(int i=1;i<=n;i++)cin>>a[i];
        dp[0]=0;
        ans=1;
        for(int i=1;i<=k;i++)dp[i]=dp[i-1]+a[i];
        
        int Min=dp[k];
        for(int i=1+k;i<=n;i++){
            dp[i]=dp[i-1]+a[i];
            int tmp=dp[i]-dp[i-k];
            if(tmp<Min){
                Min=tmp;
                ans=i-k+1;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}
原文地址:https://www.cnblogs.com/pk28/p/5936249.html