kickstart_Round C 2020

题目链接

做题情况：100% + 100% + 50% + 0%， 43分， 1935名

第一题：m序列的个数（连续m个递减且最后一个为1），n<2e5

思路：用dp记录递减序列的长度

#include<bits/stdc++.h>
using namespace std;

const int maxn = 2e5 +10;
const int INF = 0x3f3f3f3f;
int n, k, a[maxn], dp[maxn];

int main()
{
    int T, kase = 0;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &n, &k);
        for(int i = 0;i < n;i++)  scanf("%d", &a[i]);
        for(int i = 0;i < n;i++) dp[i] = -INF;
        for(int i = n-1;i >= 0;i--)
        {
            if(a[i] == a[i+1]+1)  dp[i] = dp[i+1]+1;
            if(a[i] == 1)  dp[i] = 1;
        }
        //for(int i = 0;i < n;i++)  printf("%d ", dp[i]);
        //printf("
");
        int ans = 0;
        for(int i = 0;i < n;i++)
            if(dp[i] == k)  ans++;
        printf("Case #%d: %d
", ++kase, ans);
    }
    return 0;
}

第二题：给定一个字符矩阵，判断能否从底到上堆积。（字符种类数不会超过26）

思路：统计有哪些种类的字符，dfs枚举字符的排列顺序。剪枝：每次都check(path)

#include<bits/stdc++.h>
using namespace std;

char maze[35][35];
char poly[30];
int cnt;
bool vis[30];
int r, c;
int kase;

bool check(vector<int>& path)
{
    char tmaze[35][35];
    for(int i = 0;i < r;i++)
        for(int j = 0;j < c;j++) tmaze[i][j] = maze[i][j];
    int n = path.size();
    for(int k = 0;k < n;k++)
    {
        for(int i = r-1;i >= 0;i--)
            for(int j = 0;j < c;j++)
            {
                if(tmaze[i][j] == poly[path[k]])
                {
                    if(i == r-1 || tmaze[i+1][j] == 'a')
                    {
                        tmaze[i][j] = 'a';
                    }
                    else  return false;
                }
            }
    }
    return true;
}

bool dfs(int cur, vector<int>& path)  // 考察到第cur个
{
    if(cur == cnt)
    {
        if(check(path))
        {
            printf("Case #%d: ", ++kase);
            for(int i = 0;i < cnt;i++)
            {
                printf("%c", poly[path[i]]);
            }
            printf("
");
            return true;
        }
        else  return false;
    }
    if(!check(path))  return false;
    for(int i = 0;i < cnt;i++)
    {
        if(!vis[i])
        {
            vis[i] = true;
            path.push_back(i);
            if(dfs(cur+1, path)) return true;
            vis[i] = false;
            path.pop_back();
        }
    }
    return false;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        //int r, c;
        scanf("%d%d", &r, &c);
        for(int i = 0;i < r;i++)
            scanf("%s", maze[i]);
        set<int>s;
        s.clear();
        cnt = 0;
        for(int i = 0;i < r;i++)
            for(int j = 0;j < c;j++)
            {
                char c = maze[i][j];
                if(s.count(c) == 0)
                {
                    s.insert(c);
                    poly[cnt++] = c;
                }
            }

        for(int i = 0;i < 30;i++)  vis[i] = false;
        vector<int>path;
        if(!dfs(0, path))
        {
            printf("Case #%d: -1
", ++kase);
        }
    }
    return 0;
}

第三题：求完美子序列（和为完全平方数的序列）的个数，N<1e5

思路：把前缀和放到map中，枚举每个位置，再枚举可能出现的平方数。

#include<bits/stdc++.h>
using namespace std;
using LL = long long;
constexpr int maxn = -1;
void work(){
    int N;
    cin >> N;
    int S = 0, MS = 0;
    unordered_map<int, int> mp;
    mp[S] += 1;
    LL ans = 0;
    for(int i = 1, A; i <= N; i += 1){
        cin >> A;
        S += A;
        for(int j = 0;; j += 1){
            int T = S - j * j;
            if(T < MS) break;
            if(mp.count(T)) ans += mp[T];
        }
        MS = min(MS, S);
        mp[S] += 1;
    }
    cout << ans << "
";
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int T;
    cin >> T;
    for(int t = 1; t <= T; t += 1){
        cout << "Case #" << t << ": ";
        work();
    }
    return 0;
}

第四题：一个序列，两种操作：查询和修改

思路：一看就是线段树，但是不会做。（这里记录一下大佬的代码，有空学习一下）

#include<bits/stdc++.h>
using namespace std;
using LL = long long;
constexpr int maxn = 200000;
#define ls (v << 1)
#define rs (ls | 1)
#define tm ((tl + tr) >> 1)
LL A[maxn + 1];
struct Node{
    LL ans, sum, size;
    Node operator + (const Node& node)const{
        Node res;
        res.size = size + node.size;
        if(size & 1) res.sum = sum - node.sum;
        else res.sum = sum + node.sum;
        if(size & 1) res.ans = ans - node.ans - size * node.sum;
        else res.ans = ans + node.ans + size * node.sum;
        return res; 
    }
}T[maxn << 2];
void build(int v, int tl, int tr){
    if(tl == tr){
        T[v].size = 1;
        T[v].ans = T[v].sum = A[tm];
    }
    else{
        build(ls, tl, tm);
        build(rs, tm + 1, tr);
        T[v] = T[ls] + T[rs];
    }
}
void update(int v, int tl, int tr, int x, int V){
    if(tl == tr)
        T[v].ans = T[v].sum = V;
    else{
        if(x <= tm) update(ls, tl, tm, x, V);
        else update(rs, tm + 1, tr, x, V);
        T[v] = T[ls] + T[rs];
    }
}
Node query(int v, int tl, int tr, int L, int R){
    if(tl >= L and tr <= R) return T[v];
    if(R <= tm) return query(ls, tl, tm, L, R);
    if(L > tm) return query(rs, tm + 1, tr, L, R);
    return query(ls, tl, tm, L, R) + query(rs, tm + 1, tr, L, R);
}
void work(){
   int N, Q;
   cin >> N >> Q;
   for(int i = 1; i <= N; i += 1) cin >> A[i];
   build(1, 1, N);
   LL ans = 0;
   for(int i = 1; i <= Q; i += 1){
       string s;
       cin >> s;
       if(s == "U"){
           LL X, V;
           cin >> X >> V;
           update(1, 1, N, X, V);
       }
       else{
           int L, R;
           cin >> L >> R;
           //cout << query(1, 1, N, L, R).ans << "
";
           ans += query(1, 1, N, L, R).ans;
       }
   }
   cout << ans << "
";
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int T;
    cin >> T;
    for(int t = 1; t <= T; t += 1){
        cout << "Case #" << t << ": ";
        work();
    }
    return 0;
}