Codeforces Edu Round 59 A-D

A. Digits Sequence Dividing

注意特殊情况，在\(n = 2\)时除非\(str[1] >= str[2]\)，否则可以把第一个数划分下来，剩下的数直接当成一组，一定满足条件。

#include <cstdio>
#include <iostream>
#include <string>
using namespace std;
const int N = 310;
int n;
char s[N];
int main(){
    int T; scanf("%d", &T);
    while(T--){
        scanf("%d%s", &n, s + 1);
        if(n == 2){
            if(s[1] >= s[2]) puts("NO");
            else printf("YES\n2\n%c  %c\n", s[1], s[2]);
        }else{

            printf("YES\n2\n%c ", s[1]);
            for(int i = 2; i <= n; i++) putchar(s[i]);
            puts("");
        }
    }
    return 0;
}

B. Digital root

通过打表找规律发现的…看了题解，证明还是很NB的...

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long LL;
int main(){
    int T; scanf("%d", &T);
    while(T--){
        LL k, x; scanf("%lld%lld", &k, &x);
        printf("%lld\n", (k - 1) * 9 + x);
    }
    return 0;
}

C. Brutality

用堆维护连续子段最大和即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long LL;
const int N = 200010;
int n, k, a[N];
LL ans = 0;
char s[N];
priority_queue<int, vector<int>, greater<int> > q;
int main(){
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; i++) scanf("%d", a + i);
    scanf("%s", s + 1);
    for(int i = 1; i <= n; i++){
        if(s[i] != s[i - 1]){
            while(!q.empty()) ans += q.top(), q.pop();
            q.push(a[i]);
        }else{
            q.push(a[i]);
        }
        while(q.size() > k) q.pop();
    }
    while(!q.empty()) ans += q.top(), q.pop();
    printf("%lld\n", ans);
    return 0;
}

D. Compression

实质上是把这图压缩到最小的点阵图，用\(bitset\)优化复杂度，暴力水过。

#include <cstdio>
#include <iostream>
#include <bitset>
using namespace std;
const int N = 5210;
bitset<N> a[N];
int n;
//密集恐惧症
int main(){
    ios::sync_with_stdio(false);
    cin >> n;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j += 4){
            char x; cin >> x;
          	//16进制转2进制
            if(x == '1') a[i][j + 3] = 1;
            else if(x == '2') a[i][j + 2] = 1;
            else if(x == '3') a[i][j + 2] = a[i][j + 3] = 1;
            else if(x == '4') a[i][j + 1] = 1;
            else if(x == '5') a[i][j + 1] = a[i][j + 3] = 1;
            else if(x == '6') a[i][j + 2] = a[i][j + 3] = 1;
            else if(x == '7') a[i][j + 1] = a[i][j + 2] = a[i][j + 3] = 1;
            else if(x == '8') a[i][j] = 1;
            else if(x == '9') a[i][j] = a[i][j + 3] = 1;
            else if(x == 'A') a[i][j] = a[i][j + 2] = 1;
            else if(x == 'B') a[i][j] = a[i][j + 2] = a[i][j + 3] = 1;
            else if(x == 'C') a[i][j] = a[i][j + 1] = 1;
            else if(x == 'D') a[i][j] = a[i][j + 1] = a[i][j + 3] = 1;
            else if(x == 'E') a[i][j] = a[i][j + 1] = a[i][j + 2] = 1;
            else if(x == 'F') a[i][j] = a[i][j + 1] = a[i][j + 2] = a[i][j + 3] = 1;
        }
    }
    for(int x = n; x >= 2; x--){
        if(n % x) continue;
        bool ep = true;
        for(int i = 1; i <= n; i += x){
            for(int j = i + 1; j < i + x; j++){
                ep = ep && (a[j] == a[i]);
                if(!ep) break;
            }
            if(!ep) break;
            for(int j = 1; j <= n; j += x){
                for(int k = j + 1; k < j + x; k++){
                    ep = ep && (a[i][k] == a[i][k - 1]);
                    if(!ep) break;
                }
                if(!ep) break;
            }
            if(!ep) break;
        }
        if(ep) { cout << x; return 0; }
    }
    cout << 1;
    return 0;
}