[hdu3486]rmq+枚举优化

题意：给n个数，求最小的段数，使得每一段的最大值之和大于给定的k。每一段的长度相等，最后若干个丢掉。

思路：从小到大枚举段数，如果能o(1)时间求出每一段的和，那么总复杂度是O(n(1+1/2+1/3+...+1/n))=O(nlogn)的。但题目时限卡得比较紧，需加一点小优化，如果连续两个段数它们每一段的个数一样，那么这次只比上次需要多计算一个区间，用上一次的加上这个区间最大值得到当前分段的总和，这样能减少不少运算量。详见代码：

#pragma comment(linker, "/STACK:10240000,10240000")

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <ctime>
#include <cctype>
#include <set>
#include <bitset>
#include <functional>
#include <numeric>
#include <stdexcept>
#include <utility>

using namespace std;

#define mem0(a) memset(a, 0, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define define_m int m = (l + r) >> 1
#define rep_up0(a, b) for (int a = 0; a < (b); a++)
#define rep_up1(a, b) for (int a = 1; a <= (b); a++)
#define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
#define rep_down1(a, b) for (int a = b; a > 0; a--)
#define all(a) (a).begin(), (a).end()
#define lowbit(x) ((x) & (-(x)))
#define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
#define pchr(a) putchar(a)
#define pstr(a) printf("%s", a)
#define sstr(a) scanf("%s", a);
#define sint(a) ReadInt(a)
#define sint2(a, b) ReadInt(a);ReadInt(b)
#define sint3(a, b, c) ReadInt(a);ReadInt(b);ReadInt(c)
#define pint(a) WriteInt(a)
#define if_else(a, b, c) if (a) { b; } else { c; }
#define if_than(a, b) if (a) { b; }
#define test_print1(a) cout << "var1 = " << a << endl
#define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
#define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = b" << ", var3 = " << c << endl

typedef double db;
typedef long long LL;
typedef pair<int, int> pii;
typedef multiset<int> msi;
typedef set<int> si;
typedef vector<int> vi;
typedef map<int, int> mii;

const int dx[8] = {0, 0, -1, 1};
const int dy[8] = {-1, 1, 0, 0};
const int maxn = 4e5 + 7;
const int maxm = 1e3 + 7;
const int maxv = 1e7 + 7;
const int max_val = 1e6 + 7;
const int MD = 22;
const int INF = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-10;

template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
template<class T>void ReadInt(T &x){char c=getchar();while(!isdigit(c))c=getchar();x=0;while(isdigit(c)){x=x*10+c-'0';c=getchar();}}
template<class T>void WriteInt(T i) {int p=0;static int b[20];if(i == 0) b[p++] = 0;else while(i){b[p++]=i%10;i/=10;}for(int j=p-1;j>=0;j--)pchr('0'+b[j]);}
template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
template<class T>T condition(bool f, T a, T b){return f?a:b;}
template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
int make_id(int x, int y, int n) { return x * n + y; }

int f[maxn][20], t[maxn], a[maxn], sum[maxn];
int n;
void RMQ_Init() {
    rep_up0(i, n) f[i][0] = a[i];
    rep_up1(j, 18) {
        for (int i = 0; i + (1 << j) - 1 < n; i++) {
            f[i][j] = max(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
        }
    }
}
int RMQ(int L, int R) {
    int p = t[R - L + 1];
    return max(f[L][p], f[R - (1 << p) + 1][p]);
}
LL getSum(int t, int x) {
    LL sum = 0;
    rep_up0(i, t) {
        sum += RMQ(i * x, i * x + x - 1);
    }
    return sum;
}
int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int k;
    rep_up1(i, 18) {
        for (int j = (1 << (i - 1)) + 1; j <= (1 << i); j++) t[j] = i - 1;
    }
    while (cin >> n >> k, n >= 0 || k >= 0) {
        rep_up0(i, n) {
            sint(a[i]);
            if (i) sum[i] = sum[i - 1] + a[i];
            else sum[i] = a[i];
        }
        int ans = -1, last_sum = 0;
        RMQ_Init();
        for (int i = 1; i <= n; i++) {
            int x = n / i;
            LL sum = 0;
            if (i > 1 && n / (i - 1) == x) sum = last_sum + RMQ(x * (i - 1), x * i - 1);
            else sum = getSum(i, x);
            if (sum > k) {
                ans = i;
                break;
            }
            last_sum = sum;
        }
        cout << ans << endl;
    }
    return 0;
}