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

 template<class T>T gcd(T a, T b){return b==?a:gcd(b,a%b);}
 template<class T>void ReadInt(T &x){char c=getchar();while(!isdigit(c))c=getchar();x=;while(isdigit(c)){x=x*+c-'';c=getchar();}}
 template<class T>void WriteInt(T i) {int p=;static int b[];if(i == ) b[p++] = ;else while(i){b[p++]=i%;i/=;}for(int j=p-;j>=;j--)pchr(''+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][], t[maxn], a[maxn], sum[maxn];
 int n;
 void RMQ_Init() {
     rep_up0(i, n) f[i][] = a[i];
     rep_up1(j, ) {
         for (int i = ; i + ( << j) -  < n; i++) {
             f[i][j] = max(f[i][j - ], f[i + ( << (j - ))][j - ]);
         }
     }
 }
 int RMQ(int L, int R) {
     int p = t[R - L + ];
     return max(f[L][p], f[R - ( << p) + ][p]);
 }
 LL getSum(int t, int x) {
     LL sum = ;
     rep_up0(i, t) {
         sum += RMQ(i * x, i * x + x - );
     }
     return sum;
 }
 int main() {
     //freopen("in.txt", "r", stdin);
     //freopen("out.txt", "w", stdout);
     int k;
     rep_up1(i, ) {
         for (int j = ( << (i - )) + ; j <= ( << i); j++) t[j] = i - ;
     }
     while (cin >> n >> k, n >=  || k >= ) {
         rep_up0(i, n) {
             sint(a[i]);
             if (i) sum[i] = sum[i - ] + a[i];
             else sum[i] = a[i];
         }
         int ans = -, last_sum = ;
         RMQ_Init();
         for (int i = ; i <= n; i++) {
             int x = n / i;
             LL sum = ;
             if (i >  && n / (i - ) == x) sum = last_sum + RMQ(x * (i - ), x * i - );
             else sum = getSum(i, x);
             if (sum > k) {
                 ans = i;
                 break;
             }
             last_sum = sum;
         }
         cout << ans << endl;
     }
     return ;
 }