hihocoder [Offer收割]编程练习赛24

比赛https://hihocoder.com/contest/offers24/problems

应该能拿到个鼠标垫

舒服了

这个oj体验好差，每次都要先因为选错语言CE一发，而且刷新之后还是默认GCC

A ⼩Hi的钟表

算秒数，算角度，注意180以内，所以大于180之后拿360减

#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;
int h, m, s, t;

int main(){
    //freopen("in.txt", "r", stdin);
    int _;
    scanf("%d", &_);
    for (; _--;){
        scanf("%d%d%d%d", &h, &m, &s, &t);
        t += h * 3600 + m * 60 + s;
        t %= 3600 * 12;
        m = t % 3600;
        double ans = fabs(1.0 * t / 120 - 1.0 * m / 10); 
        if (ans > 180) ans = 360 - ans; 
        //printf("t = %d\n", t);
        //double ans = 1.0 * t / 120;
        printf("%.4lf\n", ans);
    }

    return 0;
} 

B 影院座位

没想到这题居然是最难的一题，卡了一万年

hint: 中位数，前缀和

#include <cstdio>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAXN = 1e6 + 5;
int a[MAXN];
int Getc[MAXN];
int Sum[MAXN];
int main(){
    //freopen("in.txt", "r", stdin);
    int N, M;
    cin >> N >> M;
    for(int i = 1; i <= N; i++) {
        scanf("%d", Getc + i);
    }
    int cnt = 0;
    for(int i = 1; i <= N; i++) {
        if(Getc[i] == 0) {
            a[cnt++] = i;
        }
    }
    for(int i = 0; i < cnt; i++) {
        a[cnt + i] = a[i] + N;
    }
    Sum[0] = a[0];
    for(int i = 1; i < cnt * 2; i++) {
        Sum[i] = Sum[i - 1] + a[i];
    }
    if(cnt < M) {
        printf("-1\n");
        return 0;
    }
    int ans = a[M / 2] * (M / 2) - Sum[M / 2 - 1] + Sum[M - 1] - Sum[M / 2] - (M - 1 - M / 2) * a[M / 2];
    for(int i = 1; i < cnt; i++) {
        ans = min(ans, a[(i + i + M) / 2] * (M / 2) - Sum[(i + i + M) / 2 - 1] + Sum[i - 1] + Sum[M + i - 1] - Sum[(i + i + M) / 2] - (M - 1 - M / 2) * a[(i + i + M) / 2]);
    }
    printf("%d\n", ans);
    return 0;
}

C 同步H公司的终端

树形DP，统计每个点到叶子的最大距离，让每个子树都是一个“相等”、“平衡”状态

尽量修改靠上的节点，这样最后修改结果最小

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 1e5 + 7;
LL a[N], ans;
int n;
bool NotRoot[N];
int FindRoot(){
    for (int i = 1; i <= n; i++) 
        if (!NotRoot[i]) return i;
    return -1; 
} 

struct Edge{
    int from, to, nxt;
    Edge(){}
    Edge(int u, int v, int n):from(u), to(v), nxt(n){}
} edges[N * 2];
int E, head[N]; 
LL dist[N];
inline void AddEdge(int f, int t){
    edges[++E] = Edge(f, t, head[f]);
    head[f] = E;
}
inline void Init(){
    E = -1;
    for (int i = 0; i <= n; i++) head[i] = -1;
    for (int i = 0; i <= n; i++) dist[i] = 0;
}

int dfs(int u, int pre){
    //printf("dfs(%d, %d)\n", u, pre);
    LL Max = 0;
    for (int i = head[u]; i != -1; i = edges[i].nxt){
        Edge &e = edges[i];
        dfs(e.to, u);
        Max = max(Max, dist[e.to]);
    }
    dist[u] = Max + a[u];
    for (int i = head[u]; i != -1; i = edges[i].nxt){
        Edge &e = edges[i];
        ans += Max - dist[e.to];
    }
}

int main(){
    //freopen("in.txt", "r", stdin);
    int u, v;
    for (; ~scanf("%d", &n);){
        a[0] = 0;
        for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
        Init();
        for (int i = 1; i < n; i++) {
            scanf("%d%d", &u, &v);
            AddEdge(u, v);
            NotRoot[v] = 1;
        }
        int root = FindRoot();
        ans = 0;
        dfs(root, -1);
        printf("%lld\n", ans);
    }
    return 0;
}

D 大富翁

本以为是个压轴题，以为要线段树什么各种数据结构来维护

然后一看过的人多的不正常，想，先拿30分再说，然后就莫名其妙过了

标算应该不是这么粗暴的，加强数据应该还是过不了

这里就是根据题意模拟

#include <cstdio>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAXN = 1e6 + 5;
int a[MAXN];
int N;
int Find(int x) {
    if(x > N) return 0;
    return 1 + Find(x + a[x]);
}
int main(){
    cin >> N;
    for(int i = 1; i <= N; i++) {
        scanf("%d", a + i);
    }
    int Q;
    cin >> Q;
    while(Q--) {
        int x;
        scanf("%d", &x);
        if(x == 1) {
            int y;
            scanf("%d", &y);
            printf("%d\n", Find(y));
        } else {
            int u, v;
            scanf("%d%d", &u, &v);
            a[u] = v;
        }
    }
    return 0;
}

hiho一下 第165周

分隔相同字符

复杂度1e5*26*26

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1e5 + 7;
char st[N];
int Hash[222], n;

int getMax(char ch){
    int ans = 0;
    for (char i = 'a'; i <= 'z'; i++){
        if (i == ch) ans = max(ans, Hash[i]-1);
        else ans = max(ans, Hash[i]);
    }
    return ans;
}

char FindChar(char last){
    n--;
    for (char i = 'a'; i <= 'z'; i++){
        if (!Hash[i]) continue;
        if (i == last) continue;
        int Max = getMax(i);
        if (Max * 2 > n + 1) continue;
        Hash[i]--;
        return i;
    }
    return '0';
}

int main(){
    //freopen("in.txt", "r", stdin);
    scanf("%s", st);
    int len = strlen(st);
    n = len;
    memset(Hash, 0, sizeof(Hash));
    for (int i = 0; i < len; i++) Hash[st[i]]++;
    int Max = getMax('0');
    if (Max * 2 > len + 1) puts("INVALID");
    else {
        char last = ' ';
        for (int i = 0; i < len; i++){
            last = FindChar(last);
            printf("%c", last);
        }
        puts("");
    }
    return 0;
}