A-n的约数
1 #pragma warning(disable:4996) 2 #include<map> 3 #include<queue> 4 #include<string> 5 #include<vector> 6 #include<cstdio> 7 #include<cstring> 8 #include<iostream> 9 #include<algorithm> 10 using namespace std; 11 typedef long long ll; 12 13 const int maxn = 1005; 14 15 ll n, cnt; 16 int p[20] = { 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47 }; 17 18 void DFS(int k, ll sum, ll num, int Limit) { 19 if (k > 14) return; 20 if (num > cnt) cnt = num; 21 22 for (int i = 1; i <= Limit; i++) { 23 if (sum <= n / p[k]) { 24 sum *= p[k]; 25 DFS(k + 1, sum, num*(i + 1), i); 26 } 27 } 28 /* 可能会爆long long. 29 for (int i = 1; i <= Limit; i++) { 30 sum *= p[k]; 31 if (sum < n) DFS(k + 1, sum, num*(i + 1), i); 32 } 33 */ 34 } 35 36 int main() 37 { 38 int T; cin >> T; 39 while (T--) { 40 cin >> n; 41 cnt = 1; 42 DFS(0, 1, 1, 15); 43 cout << cnt << endl; 44 } 45 return 0; 46 }
B-区间的连续段
题解：用ST表记录当前起点i跳2^j次能到达的位置，然后从大到小（当前查找的区间最长）查找就行了。
感受：看到网上题解说用ST表，蒙着脑袋就开始写了，结果。。。存错对象了。死活写不出来，orzzzzz。这里ST表存的是以当前结点为左端点，不满足区间和小于或等于k的第一个出现的右端点位置（跳一次的情况，1=2^0）！
1 #pragma warning(disable:4996) 2 #include<map> 3 #include<queue> 4 #include<string> 5 #include<vector> 6 #include<cstdio> 7 #include<cstring> 8 #include<iostream> 9 #include<algorithm> 10 using namespace std; 11 typedef long long ll; 12 13 const int maxn = 1000005; 14 15 int n, m, k; 16 int b[maxn], dp[maxn][20]; 17 ll sum[maxn]; 18 19 void Inite() { 20 for (int i = 1; (1 << i) <= n; i++) { 21 for (int j = 1; j <= n; j++) dp[j][i] = dp[dp[j][i - 1]][i - 1]; 22 } 23 } 24 25 int solve(int l,int r) { 26 if (b[r] - b[l - 1] > 0) return -1; //是减去b[l-1]!!!!! 27 int ans = 1; 28 for (int i = 19; dp[l][0] <= r; i--) { 29 if (dp[l][i] && dp[l][i] <= r) { 30 l = dp[l][i]; 31 ans += (1 << i); 32 } 33 } 34 return ans; 35 } 36 37 int main() 38 { 39 while (scanf("%d%d%d", &n, &m, &k) != EOF) { 40 41 memset(b, 0, sizeof(b)); 42 memset(dp, 0, sizeof(dp)); 43 memset(sum, 0, sizeof(sum)); 44 45 for (int i = 1; i <= n; i++) { 46 int tp; 47 scanf("%d", &tp); 48 sum[i] = sum[i - 1] + tp; 49 b[i] = b[i - 1] + (tp > k); 50 } 51 52 for (int i = 1; i <= n; i++) dp[i][0] = upper_bound(sum + 1, sum + n + 1, sum[i - 1] + k) - sum; 53 Inite(); 54 55 for (int i = 1; i <= m; i++) { 56 int l, r; 57 scanf("%d%d", &l, &r); 58 int ans = solve(l, r); 59 if (ans == -1) puts("Chtholly"); 60 else printf("%d ", ans); 61 } 62 } 63 return 0; 64 }