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;
}
/*
5
1
1
1
11
6
*/D：
还不会，
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 ;
}
Tutorial CodeForces Round 289 (Div.2) (Second Winter Computer Camp Selection 2015) 题解