题目链接：点击打开链接
A：
#include <cstdio>#include <vector>#include <algorithm>#include <iostream>#include <map>#include <set>#include <queue>#include <cstring>#include <cmath>#include <string>using namespace std;int a[12][11];int main() { for (int i = 0; i <= 10; ++i) a[i][0] = a[0][i] = 1; for (int i = 1; i <= 10; ++i) { for (int j = 1; j <= 10; ++j) { a[i][j] = a[i - 1][j] + a[i][j - 1]; } } int n; scanf("%d", &n); printf("%d\n", a[n - 1][n - 1]); return 0;}B：构造#include <cstdio>#include <vector>#include <algorithm>#include <iostream>#include <map>#include <set>#include <queue>#include <cstring>#include <cmath>#include <string>using namespace std;typedef long long ll;const int N = 105;int n, m;int a[N];int main() { while (~scanf("%d %d", &n, &m)){ int mi = 1000, mx = 0; for (int i = 1; i <= n; i++){ scanf("%d", &a[i]); mi = min(mi, a[i]); mx = max(mx, a[i]); } if (mx - mi > m)puts("NO"); else { puts("YES"); for (int i = 1; i <= n; i++){ for (int j = 1; j <= mi; j++)printf("%d ", 1); a[i] -= mi; for (int j = 1; j <= a[i]; j++)printf("%d ", j); puts(""); } } } return 0;}C：模拟题意：有一个n长的严格单调递增序列，对于序列上某个点的权值就是数位和（即 123的数位和为6)
现在给出这个序列的数位和，构造一个序列使得最后一个数最小。
略麻烦，首先我们得到一个结论：对于给定的sum[i]，我们目标是构造最小的且比前面的数大的数。
一定是构造最小的数，因为大的数我们也能用sum[i]构造，倒不如构造最小的。
构造时先判断是否需要进位（即位数能不能和上一个数字一样）
然后暴力递增即可。
#include <cstdio>#include <vector>#include <algorithm>#include <iostream>#include <map>#include <set>#include <queue>#include <cstring>#include <cmath>#include <string>using namespace std;typedef long long ll;const int N = 305;int n;int Ni(vector<int>&x){//返回最低位且不为9的位置，若全为9返回-1 for (int i = 0; i < x.size(); i++)if (x[i] != 9)return i; return -1;}void add(vector<int>&x){//让x的数位和增加1 int ni = Ni(x); if (ni == -1){//全为9时想要让数位和增加1只能在最高位添一个1 x.push_back(1); } else { x[ni]++; }}void hehe(vector<int>&now, vector<int>&old, int sum, int l){//当位数比上一个数字要大时，我们先构造一个100···这样的序列，且使得这个序列位数最少且全为9时的数位和>=sum，然后暴力增加数位和即可 int dig = old.size() + 1; while (dig * 9 < sum){ dig++; } dig--; now.clear(); while (dig--)now.push_back(0); now.push_back(1); sum--; while (sum--)add(now);}void work(vector<int>&now, vector<int>&old, int sum,int l){ if (l < sum){//如果数位和已经比上一个数大就直接从上一个数开始增加数位和 now = old; sum -= l; while (sum--)add(now); return; } else { if (Ni(old) == -1 || old.size() * 9 < sum){ hehe(now, old, sum, l); return; } now = old; int s = 0; int find = -1; for (int i = now.size()-1; i >= 0; i--){//如果我们把上一个数的某一位 find 增加1，然后把个位到find位全变0，能使得数位和<=sum 那么我们就不必增加位数，否则还是要增加位数 s += now[i]; if (now[i] < 9 && s + 1 <= sum){ find = i; } } if (find!=-1){ now[find]++; for (int i = 0; i < find; i++)now[i] = 0; int ssum = sum; for (int i = 0; i < now.size(); i++)ssum -= now[i]; if (ssum >= 0){ while (ssum--){ add(now); } for (int i = now.size() - 1; i >= 0; i--)if (now[i]>old[i]) return; else if (now[i] < old[i])break; } } hehe(now, old, sum, l); }}vector<int>u[2];int main() { while (~scanf("%d", &n)){ u[0].clear(); u[1].clear(); u[1].push_back(0); int cur = 0, last = 1; int sum, l = 0; while (n--){ scanf("%d", &sum); work(u[cur], u[last], sum, l); for (int i = u[cur].size() - 1; i >= 0; i--)putchar('0' + u[cur][i]); puts(""); l = sum; cur ^= 1; last ^= 1; } } return 0;}/*5111116*/D：import java.util.*;public class D { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int m = in.nextInt(); long[][] w = new long[n][m]; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { w[i][j] = in.nextLong(); } } // check a k constraint // delta for ai - a0 in each j boolean possible = true; Long k = null; for (int i = 1; i < n; i++) { Set<Long> deltas = new TreeSet<Long>(); for (int j = 0; j < m; j++) { deltas.add(w[i][j] - w[0][j]); } if (deltas.size() > 2) { possible = false; break; } else if (deltas.size() == 2) { Iterator<Long> iterator = deltas.iterator(); long one = iterator.next(); long two = iterator.next(); if (one * two > 0) { possible = false; break; } else { long possibleK = Math.abs(one - two); if (k == null) { k = possibleK; } else if (k != possibleK) { possible = false; break; } } } } // check b k constraint // delta for bj - b0 in each i for (int j = 1; j < m; j++) { Set<Long> deltas = new TreeSet<Long>(); for (int i = 0; i < n; i++) { deltas.add(w[i][j] - w[i][0]); } if (deltas.size() > 2) { possible = false; break; } else if (deltas.size() == 2) { Iterator<Long> iterator = deltas.iterator(); long one = iterator.next(); long two = iterator.next(); if (one * two > 0) { possible = false; break; } else { long possibleK = Math.abs(one - two); if (k == null) { k = possibleK; } else if (k != possibleK) { possible = false; break; } } } } // double check k if (k != null && possible) { for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (w[i][j] >= k) { possible = false; } } } } // if (possible) { System.out.println("YES"); if (k == null) { k = 10000000000000L; } System.out.println(k); long a0 = k; StringBuilder aString = new StringBuilder(); aString.append(a0); for (int i = 1; i < n; i++) { aString.append(' ').append(a0 + (w[i][0] - w[0][0])); } System.out.println(aString.toString()); long b0 = k + w[0][0]; StringBuilder bString = new StringBuilder(); bString.append(b0); for (int j = 1; j < m; j++) { bString.append(' ').append(b0 + (w[0][j] - w[0][0])); } System.out.println(bString.toString()); } else { System.out.println("NO"); } }}E：
#include <cstdio>#include <vector>#include <algorithm>#include <iostream>#include <map>#include <set>#include <queue>#include <cstring>#include <cmath>#include <string>using namespace std;const int MAX_N = 500007;int bit[MAX_N << 1], len;int sum(int i) { int s = 0; while (i > 0) { s += bit[i]; i -= i & -i; } return s;}void add(int i, int x) { while (i <= len) { bit[i] += x; i += i & -i; }}char str[MAX_N];bool is(char ch) { return ch == 'A' || ch == 'E' || ch == 'I' || ch == 'O' || ch == 'U' || ch == 'Y';}long long z[MAX_N];long long zz[MAX_N];int main() { scanf("%s", str + 1); len = strlen(str + 1); for (int i = 1; str[i]; ++i) { if (is(str[i])) z[i] = z[i - 1] + 1; else z[i] = z[i - 1]; } for (int i = 1; i <= len; ++i) { zz[i] = zz[i - 1] + z[i]; } double ans = 0; for (int i = 0; i <= len; ++i) { if (len - (i + 1) >= 0) ans += (zz[len] - zz[len - (i + 1)] - (zz[i])) * 1. / (i + 1); } printf("%f\n", ans); return 0;}F：dp[l][r]表示 区间[l,r] 构成一棵树的方法数。
对于一个区间[l, r] 构成一棵树，则点l一定是根，然后枚举2个区间相乘即可
dp[l][r] = dp[l+1][i] * dp[i+1][r] ( i = [l+1, r] )
当然 a[i+1] > a[l+1] ，这样才会满足题目中的暴力代码。。==
import java.io.PrintWriter;import java.text.DecimalFormat;import java.util.ArrayList;import java.util.Arrays;import java.util.Collections;import java.util.Comparator;import java.util.HashMap;import java.util.Iterator;import java.util.LinkedList;import java.util.Map;import java.util.PriorityQueue;import java.util.Scanner;import java.util.Stack;import java.util.TreeMap;import java.util.TreeSet;import java.util.Queue;public class Main { int n; int[] a = new int[N]; long[][] dp = new long[N][N]; long dfs(int l, int r){ if(dp[l][r] != -1)return dp[l][r]; if(l >= r)return dp[l][r] = 1; long ans = 0; for(int i = l+1; i <= r; i++) if(i == r || a[i + 1] > a[l + 1]) ans = (dfs(l+1, i) * dfs(i, r)+ans)%mod; return dp[l][r] = ans; } void work() { while(cin.hasNext()){ n = cin.nextInt(); for(int i = 1; i <= n; i++)a[i] = cin.nextInt(); for(int i = 1; i <= n; i++){ for(int j = i; j <= n; j++)dp[i][j] = -1; } System.out.println(dfs(1, n)); } } Main() { cin = new Scanner(System.in); out = new PrintWriter(System.out); } public static void main(String[] args) { Main e = new Main(); e.work(); out.close(); } public Scanner cin; public static PrintWriter out; static int N = 505; DecimalFormat df=new DecimalFormat("0.0000"); static int mod = 1000000007 ;}