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暑假N天乐【比赛篇】 —— 2019牛客暑期多校训练营(第六场)

作者:互联网

疯狂摸鱼ing...不想写了。

以下题解包括:\(A \ \ \ B \ \ \ C \ \ \ D \ \ \ E \ \ \ G \ \ \ J\)

比赛地址: https://ac.nowcoder.com/acm/contest/886#question

【A】 Garbage Classification 水题

签到题,不说了。

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;

const int maxn = 2000+5;

map<char, char> mp;
char a[maxn], b[maxn];

int main() {
    int t, cas = 1;
    scanf("%d", &t);
    while(t--) {
        mp.clear();
        scanf("%s%s", a, b);
        for(int i = 0; i < 26; i++) {
            mp[i+'a'] = b[i];
        }
        int n = strlen(a);
        int H = 0, W = 0, D = 0;
        for(int i = 0; i < n; i++) {
            if(mp[a[i]] == 'h') {
                H ++;
            }
            else if(mp[a[i]] == 'd') {
                D ++;
            }
            else {
                W ++;
            }
        }
        printf("Case #%d: ", cas++);
        if(H*4 >= n) {
            printf("Harmful\n");
        }
        else if(H*10 <= n) {
            printf("Recyclable\n");
        }
        else if(D >= 2*W){
            printf("Dry\n");
        }
        else {
            printf("Wet\n");
        }
    }
    return 0;
}

【B】 Shorten IPv6 Address 模拟

给定一个长度 128 位的 01 字符串,按照规则进行变换成16进制的 \(ipv6\) 地址,输出变换后字典序最小的字符串。

纯大模拟,虽然 \(python\) 很好写,但我不会啊。代码是现场敲的,极丑将就看吧。

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;

const int maxn = 1e3+5;

char s[maxn];
char str[50][50];
char str_[50][50];
struct node {
    int l, r;
    bool operator < (const node& a) const{
        if((r-l+1) == (a.r-a.l+1)) {
            return l > a.l;
        }
        return (r-l+1) > (a.r-a.l+1);
    }
}p[50];

int main() {
    int t, cas=1;
    scanf("%d", &t);
    while(t--) {
        memset(str, '\0', sizeof(str));
        memset(str_, '\0', sizeof(str_));
        memset(s, '\0', sizeof(s));
        int cnt = 0;
        scanf("%s", s);
        for(int i = 0; i < 128; i+=16) {
            char temp[50];
            int zzz = 0;
            for(int j = i; j < i+16; j+=4) {
                int sum = 0;
                int x1 = s[j] - '0';
                int x2 = s[j+1] - '0';
                int x3 = s[j+2] - '0';
                int x4 = s[j+3] - '0';
                sum = sum + (x1*8 + x2*4 + x3*2 + x4);

                if(sum >= 10) {
                    temp[zzz++] = 'a'+sum-10;
                }
                else {
                    temp[zzz++] = '0'+sum;
                }
            }
            temp[zzz] = '\0';
            strcpy(str[cnt++], temp);
        }
        for(int i = 0; i < cnt; i++) {
            int flag = 1;
            for(int j = 0; j < 4; j++) {
                if(str[i][j] != '0') {
                    flag = 0;
                }
            }
            // cout << flag << endl;
            if(flag) {  // 全 0
                str_[i][0] = '0';
            }
            else {  // 去前导 0
                int j;
                for(j = 0; j < 4; j++) {
                    if(str[i][j] != '0') {
                        break;
                    }
                }
                for(int k = j; k < 4; k++) {
                    str_[i][k-j] = str[i][k];
                }
                str_[i][4-j] = '\0';
            }
        }
        int tot = 0;
        int j = 0;
        for(int i = 0; i < cnt; i++) {
            if(strcmp(str_[i], "0") == 0) {
                for(j = i+1; j < cnt; j++) {
                    if(strcmp(str_[j], "0") != 0) {
                        break;
                    }
                }
                p[tot].l = i;
                p[tot++].r = j-1;
            }
        }
        sort(p, p+tot);
        p[tot].l = inf;
        p[tot++].r = 0;
        char ans[maxn] = {'\0'};
        int MIN = inf;
        for(int k = 0; k < tot; k++) {
            char tep[maxn] = {'\0'};
            int len = p[k].r - p[k].l + 1;
            int ok = 0;
            int res = 0;
            if(len >= 2) {
                ok = 1;
            }
            for(int i = 0; i < cnt; i++) {
                if(ok == 0) {
                    int m = strlen(str_[i]);
                    for(int j = 0; j < m; j++) {
                        tep[res++] = str_[i][j];
                    }
                    if(i==cnt-1) {
                        tep[res++] = '\0';
                    }
                    else {
                        tep[res++] = ':';
                    }
                }
                else {
                    if(p[k].l == i) {
                        if(i == 0) {
                            tep[res++] = ':';
                            tep[res++] = ':';
                        }
                        else {
                            tep[res++] = ':';
                        }
                        i = p[k].r;
                        if(i == cnt-1) {
                            tep[res++] = '\0';
                        }
                        continue;
                    }
                    int m = strlen(str_[i]);
                    for(int j = 0; j < m; j++) {
                        tep[res++] = str_[i][j];
                    }
                    if(i==cnt-1) {
                        tep[res++] = '\0';
                    }
                    else {
                        tep[res++] = ':';
                    }
                }
            }
            if(res < MIN) {
                strcpy(ans, tep);
                MIN = res;
            }
            else if(res == MIN) {
                if(strcmp(ans, tep) > 0) {
                    strcpy(ans, tep);
                }
            }
        }
        printf("Case #%d: ", cas++);
        printf("%s\n", ans);
    }   
    return 0;
}

【C】 Palindrome Mouse 回文自动机

回文自动机套模板。

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;

const int maxn = 100005;
const int N = 26;

ll ans, res;

struct Palindromic_Tree {
    int nxt[maxn][N];   // next指针,next指针和字典树类似,指向的串为当前串两端加上同一个字符构成
    int fail[maxn];     // fail指针,失配后跳转到fail指针指向的节点
    ll cnt[maxn];       // 表示节点i表示的本质不同的串的个数(建树时求出的不是完全的,最后count()函数跑一遍以后才是正确的)
    int num[maxn];      // 表示以节点i表示的最长回文串的最右端点为回文串结尾的回文串个数
    int len[maxn];      // len[i]表示节点i表示的回文串的长度(一个节点表示一个回文串)
    int S[maxn];        // 存放添加的字符
    int last;           // 指向新添加一个字母后所形成的最长回文串表示的节点。
    int n;              // 表示添加的字符个数。
    int p;              // 表示添加的节点个数。
    int vis[maxn];
    int newnode(int l) {// 新建节点
        for(int i = 0; i < N; ++i) 
            nxt[p][i] = 0;
        cnt[p] = 0;
        num[p] = 0;
        len[p] = l;
        vis[p] = 0;
        return p++;
    }
    void init() {   // 初始化
        p = 0;
        newnode(0);
        newnode(-1);
        last = 0;
        n = 0;
        S[n] = -1;  // 开头放一个字符集中没有的字符,减少特判
        fail[0] = 1;
    }
    int get_fail(int x) {   // 和KMP一样,失配后找一个尽量最长的
        while(S[n-len[x]-1] != S[n]) x = fail[x];
        return x;
    }
    void add(int c) {
        c -= 'a';
        S[++n] = c;
        int cur = get_fail(last);   // 通过上一个回文串找这个回文串的匹配位置
        if(!nxt[cur][c]) {     // 如果这个回文串没有出现过,说明出现了一个新的本质不同的回文串
            int now = newnode(len[cur]+2);     // 新建节点
            fail[now] = nxt[get_fail(fail[cur])][c];    // 和AC自动机一样建立fail指针,以便失配后跳转
            nxt[cur][c] = now;
            num[now] = num[fail[now]] + 1;
        }
        last = nxt[cur][c];
        cnt[last] ++;
    }
    void dfs(int x) {
        stack<int> s;
        int y = fail[x];
        while(y > 1 && vis[y] == 0) {
            res ++;
            s.push(y);
            vis[y] = 1;
            y = fail[y];
        }
        if(x > 1) {
            ans += res;
            vis[x] = 1;
            res ++;
        }
        for(int i = 0; i < 26; i++) {
            if(nxt[x][i]) {
                dfs(nxt[x][i]);
            }
        }
        if(x > 1) {
            vis[x] = 0;
            res --;
        }
        y = x;
        while(!s.empty()) {
            res --;
            vis[s.top()] = 0;
            s.pop();
        }
    }
}T;

int main() {
    string a;
    int t;
    int cas = 1;
    cin >> t;
    while(t--) {
        cin >> a;
        T.init();
        int len = a.size();
        for(int i = 0; i < len; i++)
            T.add(a[i]);
        res = ans = 0;
        T.dfs(0);
        T.dfs(1);
        cout << "Case #" << cas++ << ": ";
        cout << ans << endl;
    }
}

【D】 Move 思维/二分

虽然是数据水所以二分过了,但是我懒得写正解,就这样。

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 1e3+5;
 
int n, k;
int a[maxn];
int sk[maxn];
multiset<int> s;
 
int solve(int x) {
    multiset<int> temp;
    temp = s;
    for(int i = 1; i <= k; i++) {
        int sum = x;
        while(sum>0) {
            auto it = temp.upper_bound(sum);
            if(it == temp.begin()) {
                break;
            }
            it--;          
            sum -= (*it);
            temp.erase(it);
        }      
        if(temp.size() == 0) {
            return 1;
        }
    }
    return 0;
}
 
int main() {
    int t, cas = 1;
    scanf("%d", &t);
    while(t--) {
        s.clear();
        scanf("%d%d", &n, &k);
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            s.insert(a[i]);
        }
        int l = 1, r = 1000*1000+5;
        int ans = 1;
        while(l <= r) {
            int mid = (l+r) >> 1;
            int res = solve(mid);
            if(res == 1) {
                ans = mid;
                r = mid-1;
            }
            else {
                l = mid+1;
            }
        }
        for(int i = max(0, ans-100); i <= ans; i++) {
            if(solve(i)) {
                printf("Case #%d: ", cas++);
                printf("%d\n", i);
                break;
            }
        }
    }
    return 0;
}

【E】 Androgynos 构造

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;

const int maxn = 2e3+5;

int mp[maxn][maxn];
int ans[maxn];
int n;

int main() {
    int t, cas = 1;
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        printf("Case #%d: ", cas++);
        if(n % 4 != 0 && n % 4 != 1) {
            printf("No\n");
        }
        else {
            printf("Yes\n");
            int k = n / 4;
            memset(mp, 0, sizeof(mp));
            /*
                团 1  独立集 2  独立集 3  团 4
            */
            // 团 1 内部
            for(int i = 1; i <= k; i++) {
                for(int j = 1; j <= k; j++) {
                    if(i != j) {
                        mp[i][j] = mp[j][i] = 1;
                    }
                }
            }
            // 团 1 --> 独立集 2
            for(int i = 1; i <= k; i++) {
                for(int j = k+1; j <= 2*k; j++) {
                    mp[i][j] = mp[j][i] = 1;
                }
            }
            // 独立集 2 --> 独立集 3
            for(int i = k+1; i <= 2*k; i++) {
                for(int j = 2*k+1; j <= 3*k; j++) {
                    mp[i][j] = mp[j][i] = 1;
                }
            }
            // 独立集 3 --> 团 4
            for(int i = 2*k+1; i <= 3*k; i++) {
                for(int j = 3*k+1; j <= 4*k; j++) {
                    mp[i][j] = mp[j][i] = 1;
                }
            }
            // 团 4 内部
            for(int i = 3*k+1; i <= 4*k; i++) {
                for(int j = 3*k+1; j <= 4*k; j++) {
                    if(i != j) {
                        mp[i][j] = mp[j][i] = 1;
                    }
                }
            }
            // 1 2 3 4 ---> ans[i] = 2 4 1 3 
            for(int i = 1; i <= k; i++) {
                ans[i] = i+k;
                ans[i+k] = i+3*k;
                ans[i+2*k] = i;
                ans[i+3*k] = i+2*k;
            }
            // 多出来的点连向两个团 1 4 即可
            if(n % 4 == 1) {
                for(int i = 1; i <= k; i++) {
                    mp[i][n] = mp[n][i] = 1;
                }
                for(int i = 3*k+1; i <= 4*k; i++) {
                    mp[i][n] = mp[n][i] = 1;
                }
                ans[n] = n;
            }
            for(int i = 1; i <= n; i++) {
                for(int j = 1; j <= n; j++) {
                    printf("%d", mp[i][j]);
                }
                printf("\n");
            }
            for(int i = 1; i <= n; i++) {
                printf("%d%c", ans[i], i==n?'\n':' ');
            }
        }
    }
    return 0;
}

【G】 Is Today Friday? 暴力

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;

const int maxn = 1e5+5;

int n;
string str[maxn];
int ord[10];

int check(int x) {
    int year = ord[str[x][0]-'A']*1000 + ord[str[x][1]-'A']*100 + ord[str[x][2]-'A']*10 + ord[str[x][3]-'A'];
    int month = ord[str[x][5]-'A']*10 + ord[str[x][6]-'A'];
    int day = ord[str[x][8]-'A']*10 + ord[str[x][9]-'A'];
    int d_max;
    if(month==1 || month==3 || month==5 || month==7 || month==8 || month==10 || month==12) {
        d_max = 31;
    }
    else if(month==4 || month==6 || month==9 || month==11) {
        d_max = 30;
    }
    else if(month==2 && ((year%4==0 && year%100!=0) || year%400==0)) {
        d_max = 29;
    }
    else {
        d_max = 28;
    }
    if(year<1600 || year>9999 || month<1 || month>12 || day<1 || day>d_max) {
        return 0;
    }
    if(month==1 || month==2) {
        year--;
        month += 12;
    }
    int w = (day + 2*month + 3*(month+1)/5 + year + year/4 - year/100 + year/400) % 7 + 1;
    if(w == 5) {
        return 1;
    } 
    return 0;
}

int main() {
    int t, cas = 1;
    scanf("%d", &t);
    while(t--) {
        for(int i = 0; i < 10; i++) {
            ord[i] = i;
        }
        scanf("%d", &n);
        for(int i = 1; i <= n; i++) {
            cin >> str[i];
        }
        sort(str+1, str+1+n);
        n = unique(str+1, str+1+n) - (str+1);
        int ok = 0;
        do {
            ok = 1;
            for(int i = 1; i <= n; i++) {
                if(check(i) == 0) {
                    ok = 0;
                    break;
                }
            }
            if(ok) {
                break;
            }
        }while(next_permutation(ord, ord+10));
        printf("Case #%d: ", cas++);
        if(ok == 0) {
            printf("Impossible\n");
        }
        else {
            for(int i = 0; i < 10; i++) {
                printf("%d", ord[i]);
            }
            printf("\n");
        }
    }
    return 0;
}

【J】 Upgrading Technology 贪心

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
 
typedef long long ll;
const int maxn=1019;
const ll INF = 1ll*(1e9)*(1e5);
ll a[maxn][maxn],d[maxn];
ll pre[maxn][maxn];
ll P[maxn];
ll tail[maxn];
int n,m;
ll minn;
ll ans1,ans2,minj;
 
int main(){
    int T;
    scanf("%d",&T);
    int kase=1;
    while(T--){
        scanf("%d%d",&n,&m);
        memset(pre,0,sizeof(pre));
        memset(P,0,sizeof(P));
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                scanf("%lld",&a[i][j]);
            }
        }
        for(int i=1;i<=m;i++){
            scanf("%lld",&d[i]);
        }
         
        for(int i=1;i<=m;i++){
            for(int j=1;j<=n;j++){
                P[i]-=a[j][i];
            }
            P[i]=P[i-1]+P[i]+d[i];
        }
        memset(tail,0,sizeof(tail));
        //tail[n]!
        ll ans=P[m];
        for(int i=m-1;i>=0;i--){
            int flag=0;minn=-INF;
            for(int j=1;j<=n;j++){
                if(tail[j]+a[j][i+1]<=0){
                    tail[j]+=a[j][i+1];
                    P[i]-=tail[j];
                    if(minn<=tail[j]){
                        minn=tail[j];
                        minj=j;
                    }
                }
                else{
                    tail[j]=0;
                    flag=1;
                }
            }
            if(!flag){
                tail[minj]=0;
                P[i]+=minn;
            }
            ans=max(ans,P[i]);
        }
        if(ans<0) ans=0;
        printf("Case #%d: %lld\n",kase++,ans);
    }
}

标签:第六场,const,int,天乐,多校,++,maxn,str,include
来源: https://www.cnblogs.com/Decray/p/11327795.html