[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：

  1 #pragma comment(linker, "/STACK:10240000,10240000")
  2
  3 #include <iostream>
  4 #include <cstdio>
  5 #include <algorithm>
  6 #include <cstdlib>
  7 #include <cstring>
  8 #include <map>
  9 #include <queue>
 10 #include <deque>
 11 #include <cmath>
 12 #include <vector>
 13 #include <ctime>
 14 #include <cctype>
 15 #include <set>
 16 #include <bitset>
 17 #include <functional>
 18 #include <numeric>
 19 #include <stdexcept>
 20 #include <utility>
 21
 22 using namespace std;
 23
 24 #define mem0(a) memset(a, 0, sizeof(a))
 25 #define lson l, m, rt << 1
 26 #define rson m + 1, r, rt << 1 | 1
 27 #define define_m int m = (l + r) >> 1
 28 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
 29 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
 30 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
 31 #define rep_down1(a, b) for (int a = b; a > 0; a--)
 32 #define all(a) (a).begin(), (a).end()
 33 #define lowbit(x) ((x) & (-(x)))
 34 #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) {}
 35 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 36 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 37 #define pchr(a) putchar(a)
 38 #define pstr(a) printf("%s", a)
 39 #define sstr(a) scanf("%s", a);
 40 #define sint(a) ReadInt(a)
 41 #define sint2(a, b) ReadInt(a);ReadInt(b)
 42 #define sint3(a, b, c) ReadInt(a);ReadInt(b);ReadInt(c)
 43 #define pint(a) WriteInt(a)
 44 #define if_else(a, b, c) if (a) { b; } else { c; }
 45 #define if_than(a, b) if (a) { b; }
 46 #define test_print1(a) cout << "var1 = " << a << endl
 47 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
 48 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = b" << ", var3 = " << c << endl
 49
 50 typedef double db;
 51 typedef long long LL;
 52 typedef pair<int, int> pii;
 53 typedef multiset<int> msi;
 54 typedef set<int> si;
 55 typedef vector<int> vi;
 56 typedef map<int, int> mii;
 57
 58 const int dx[8] = {0, 0, -1, 1};
 59 const int dy[8] = {-1, 1, 0, 0};
 60 const int maxn = 4e5 + 7;
 61 const int maxm = 1e3 + 7;
 62 const int maxv = 1e7 + 7;
 63 const int max_val = 1e6 + 7;
 64 const int MD = 22;
 65 const int INF = 1e9 + 7;
 66 const double pi = acos(-1.0);
 67 const double eps = 1e-10;
 68
 69 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
 70 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();}}
 71 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]);}
 72 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
 73 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
 74 template<class T>T condition(bool f, T a, T b){return f?a:b;}
 75 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
 76 int make_id(int x, int y, int n) { return x * n + y; }
 77
 78 int f[maxn][20], t[maxn], a[maxn], sum[maxn];
 79 int n;
 80 void RMQ_Init() {
 81     rep_up0(i, n) f[i][0] = a[i];
 82     rep_up1(j, 18) {
 83         for (int i = 0; i + (1 << j) - 1 < n; i++) {
 84             f[i][j] = max(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
 85         }
 86     }
 87 }
 88 int RMQ(int L, int R) {
 89     int p = t[R - L + 1];
 90     return max(f[L][p], f[R - (1 << p) + 1][p]);
 91 }
 92 LL getSum(int t, int x) {
 93     LL sum = 0;
 94     rep_up0(i, t) {
 95         sum += RMQ(i * x, i * x + x - 1);
 96     }
 97     return sum;
 98 }
 99 int main() {
100     //freopen("in.txt", "r", stdin);
101     //freopen("out.txt", "w", stdout);
102     int k;
103     rep_up1(i, 18) {
104         for (int j = (1 << (i - 1)) + 1; j <= (1 << i); j++) t[j] = i - 1;
105     }
106     while (cin >> n >> k, n >= 0 || k >= 0) {
107         rep_up0(i, n) {
108             sint(a[i]);
109             if (i) sum[i] = sum[i - 1] + a[i];
110             else sum[i] = a[i];
111         }
112         int ans = -1, last_sum = 0;
113         RMQ_Init();
114         for (int i = 1; i <= n; i++) {
115             int x = n / i;
116             LL sum = 0;
117             if (i > 1 && n / (i - 1) == x) sum = last_sum + RMQ(x * (i - 1), x * i - 1);
118             else sum = getSum(i, x);
119             if (sum > k) {
120                 ans = i;
121                 break;
122             }
123             last_sum = sum;
124         }
125         cout << ans << endl;
126     }
127     return 0;
128 }