Codeforces Round #637 (Div. 2) B. Nastya and Door（前綴和）

http://codeforces.com/contest/1341/problem/B

找出長度爲k的一個區間，這一段區間裏面的“峯”最多。如果有x個峯，那麼就會造成x+1個片段。問片段數最多是多少？區間的左端點是什麼？

就是前綴和，傻傻地我用暴力T了。還是沒有吃透前綴和，關鍵時刻想不出來。

#include <iostream>
#include <bitset>
#include <algorithm>
#include <cmath>
#include <ctype.h>
#include <string>
#include <cstring>
#include <cstdio>
#include <set>
#include <fstream>
#include <deque>
#include <vector>
#include <queue>
#include <map>
#include <stack>
#include <iomanip>
#define SIS std::ios::sync_with_stdio(false)
#define ll long long
#define INF 0x3f3f3f3f
const int mod = 1e9 + 7;
const double esp = 1e-5;
const double PI = 3.141592653589793238462643383279;
using namespace std;
const int N = 1e7 + 5;
const int maxn = 1 << 20;
ll powmod(ll a, ll b) { ll res = 1; a %= mod; while (b >= 0); for (; b; b >>= 1) { if (b & 1)res = res * a % mod; a = a * a % mod; }return res; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
/*void chafen(int l, int r, int k) {//差分函數
    p[l] += k;
    p[r + 1] -= k;
}*/
/*********************************************************/
int a[N];
int sum[N];
int main()
{
    int t;
    cin >> t;
    while (t--) {
        int n, k;
        cin >> n >> k;
        for (int i = 1; i <= n; i++)
            cin >> a[i];
        sum[1] = 0; sum[n] = 0; sum[0] = 0;
        for (int i = 2; i <= n-1; i++) {
            if (a[i] > a[i - 1] && a[i] > a[i + 1])
                sum[i] = 1;
            else
                sum[i] = 0;
        }
        for (int i = 1; i <= n; i++)
                sum[i] += sum[i - 1];
        int ans = 0; int maxn = -1, pos = 0;
        a[0] = a[n + 1] = INF;
        for (int i = 1; i <= n; i++) {
            if (i + k - 1 > n)
                break;
            ans = sum[i + k - 1] - sum[i-1];
            if (a[i] > a[i - 1] && a[i] > a[i + 1])
                ans--;
            if (a[i + k - 1] > a[i + k - 2] && a[i + k - 1] > a[i + k])
                ans--;
            if (ans > maxn) {
                maxn = ans;
                pos = i;
            }
        }
        cout << maxn+1 << " " << pos << endl;
    }
    return 0;
}
/*

*/