HDOJ 3486 Interviewe ST RMQ

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<cmath>
 5 using namespace std;
 6 
 7 int n, k, tot, maxi;
 8 int d[200100][20];
 9 void init()
10 {        
11     for (int j = 1; (1 << j) < n; j++){
12         int t = (1 << j) - 1;
13         for (int i = 0; i+t < n; i++){
14             d[i][j] = max(d[i][j-1], d[i+(1<<(j-1))][j-1]);
15         }
16     }
17 }
18 inline int rmq(int a, int b)
19 {
20     int l = int(log(double(b-a+1))/log(2.0));
21     return max(d[a][l], d[b+1-(1<<l)][l]);
22 }
23 /*
24 int cal(int len)　　　　　　　　//这样写就错了，比如n=3, m=2，本来应该取2段，这个会取3段，导致答案偏小
25 {
26     int sum = 0;
27     for (int i = 0; i+len-1 < n; i += len){
28         int j = i+len-1;
29         sum += rmq(i, j);
30         if (sum > k) break;
31     }
32     return sum;
33 }
34 */
35 int main()
36 {
37     while(scanf("%d %d", &n, &k) && n >= 0 && k >= 0)
38     {
39         tot = maxi = 0;
40         for (int i = 0; i < n; i++){
41             scanf("%d", &d[i][0]);
42             tot += d[i][0];
43             maxi = max(maxi, d[i][0]);
44         }
45         if (tot <= k){
46             puts("-1");
47             continue;
48         }
49         init();
50         int ans = n;
51         for (int i = max(1, k/maxi); i < n; i++){
52             //int tmp = cal(n/i);
53             int len = n/i;
54             int tmp = 0;
55             for (int j = 1; j <= i; j++){
56                 tmp += rmq((j-1)*len, j*len-1);
57                 if (tmp > k) break;
58             }
59             if (tmp > k){
60                 ans = i;
61                 break;
62             }
63         }
64         printf("%d\n", ans);
65     }
66     return 0;
67 }