Educational Codeforces Round 51 (Rated for Div. 2)

做了四个题。。

A. Vasya And Password

直接特判即可，，为啥泥萌都说难写，，，，

这个子串实际上是忽悠人的，因为每次改一个字符就可以

我靠我居然被hack了？？？？

%……&*（）（*&……好吧我把$0$从数字里扔了。

/*

*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long
#define LL long long
#define ull unsigned long long
#define rg register
#define pt(x) printf("%d ", x);
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *O = obuf;
//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
//#define OS  *O++ = ' ';
using namespace std;
//using namespace __gnu_pbds;
const int MAXN = 1e6 + , INF = 1e9 + , mod = 1e9 + ;
const double eps = 1e-;
inline int read() {
    char c = getchar(); int x = , f = ;
    while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * f;
}
int T;
char s[MAXN];
void solve() {
    int N = strlen(s + );
    int a = , b = , c = ;
    for(int i = ; i <= N; i++) {
        if(s[i] >= '' && s[i] <= '') a++;
        if(s[i] >= 'a' && s[i] <= 'z') b++;
        if(s[i] >= 'A' && s[i] <= 'Z') c++;
    }
    if(a == ) {
        if(b > ) {
            for(int i = ; i <= N; i++)
                if(s[i] >= 'a' && s[i] <= 'z') {s[i] = ''; break;}
            b--;
        } else if(c > ) {
            for(int i = ; i <= N; i++)
                if(s[i] >= 'A' && s[i] <= 'Z') {s[i] = ''; break;}
            c--;
        }
    }
    if(b == ) {
        if(a > ) {
            for(int i = ; i <= N; i++)
                if(s[i] >= '' && s[i] <= '') {s[i] = 'a'; break;}
            a--;
        } else if(c > ) {
            for(int i = ; i <= N; i++)
                if(s[i] >= 'A' && s[i] <= 'Z') {s[i] = 'a'; break;}
            c--;
        }
    }
    if(c == ) {
        if(a > ) {
            for(int i = ; i <= N; i++)
                if(s[i] >= '' && s[i] <= '') {s[i] = 'A'; break;}
            a--;
        } else if(b > ) {
            for(int i = ; i <= N; i++)
                if(s[i] >= 'a' && s[i] <= 'z') {s[i] = 'A'; break;}
            b--;
        }
    }
    printf("%s\n", s + );
}
main() {
    T = read();
    while(T--) {
        scanf("%s", s + );
        solve();
    }
    return ;
}
/*
2 2 1
1 1
2 1 1
*/

A

B. Relatively Prime Pairs

很显然，$i$和$i+1$是互质的。

做完了

/*

*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long
#define LL long long
#define ull unsigned long long
#define rg register
#define pt(x) printf("%d ", x);
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *O = obuf;
//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
//#define OS  *O++ = ' ';
using namespace std;
//using namespace __gnu_pbds;
const int MAXN = 1e6 + , INF = 1e9 + , mod = 1e9 + ;
const double eps = 1e-;
inline int read() {
    char c = getchar(); int x = , f = ;
    while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * f;
}
int l, r;
main() {
    l = read(), r = read();
    if(l == r) {puts("NO"); return ;}
    puts("YES");
    for(int i = l; i <= r - ; i += ) {
        cout << i << " " << i +  << endl;
    }
    return ;
}
/*
2 2 1
1 1
2 1 1
*/

B

C. Vasya and Multisets

显然，如果有偶数个优秀的，对半分就可以

如果有奇数个，直接拿出一个$\geqslant 3$的数加到小的里面

否则无解

/*

*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long
#define LL long long
#define ull unsigned long long
#define rg register
#define pt(x) printf("%d ", x);
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *O = obuf;
//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
//#define OS  *O++ = ' ';
using namespace std;
//using namespace __gnu_pbds;
const int MAXN = 1e6 + , INF = 1e9 + , mod = 1e9 + ;
const double eps = 1e-;
inline int read() {
    char c = getchar();
    int x = , f = ;
    while(c < '' || c > '') {
        if(c == '-') f = -;c = getchar();}while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * f;
}
int N, A[MAXN], timssssss[MAXN];
main() {
    int n=read();
    for(int i = ; i <= n; ++i) ++timssssss[A[i] = read()];
    int cnt[] = {,,,,};
    for(int i = ; i <= ; ++i)
        if(timssssss[A[i]] < ) ++cnt[timssssss[A[i]]];
        else ++cnt[];
    if(cnt[] &  && cnt[] == ) return puts("NO"),;
    puts("YES");
    int mid = cnt[]>>;
    if(cnt[]&) {
        int p = ;
        for(int i = ; i <= n; ++i)
            if(timssssss[A[i]] > ) {
                p = i;
                break;
            }
        for(int i =  ; i <= n; ++i)
            if(timssssss[A[i]] == ) {
                if(mid) putchar('A'), --mid;
                else putchar('B');
            } else if(i != p) putchar('B');
            else putchar('A');
    } else {
        for(int i = ; i <= n; ++i)
            if(timssssss[A[i]] == ) {
                if(mid) putchar('A'), --mid;
                else putchar('B');
            } else putchar('B');
    }
    return ;
}
/*
2 2 1
1 1
2 1 1
*/

C

D. Bicolorings

普及dp？？。。。

因为只有两行，考虑把列的状态记下来

$f[i][j][sta]$表示到第$i$列，有$j$个连通块的方案，当前列的状态为$sta$，就是“白白” “白黑”“黑白”“黑黑”这四种状态

转移的时候枚举上一行选了啥

/*

*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long
#define LL long long
#define ull unsigned long long
#define rg register
#define pt(x) printf("%d ", x);
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *O = obuf;
//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
//#define OS  *O++ = ' ';
using namespace std;
//using namespace __gnu_pbds;
const int MAXN = 1e6 + , INF = 1e9 + , mod = ;
const double eps = 1e-;
inline int read() {
    char c = getchar();
    int x = , f = ;
    while(c < '' || c > '') {
        if(c == '-') f = -;c = getchar();}while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * f;
}
int N, K;
int f[][][];
// 0 �װ�
// 1 �׺�
// 2 �ڰ�
// 3 �ں�
main() {
    N = read(); K = read();
    f[][][] = ;
    f[][][] = ;
    f[][][] = ;
    f[][][] = ;
    for(int i = ; i <= N; i++) {//��i���
        for(int j = ; j <= K; j++) {//��j���� 

            (f[i][j][] += f[i - ][j][]) %= mod;
            (f[i][j][] += f[i - ][j][]) %= mod;
            (f[i][j][] += f[i - ][j][]) %= mod;
            (f[i][j][] += f[i - ][j - ][]) %= mod;

            (f[i][j][] += f[i - ][j - ][]) %= mod;
            (f[i][j][] += f[i - ][j][]) %= mod;
            (f[i][j][] += f[i - ][j - ][]) %= mod;
            (f[i][j][] += f[i - ][j - ][]) %= mod;            

            (f[i][j][] += f[i - ][j - ][]) %= mod;
            (f[i][j][] += f[i - ][j - ][]) %= mod;
            (f[i][j][] += f[i - ][j][]) %= mod;
            (f[i][j][] += f[i - ][j - ][]) %= mod;

            (f[i][j][] += f[i - ][j - ][]) %= mod;
            (f[i][j][] += f[i - ][j][]) %= mod;
            (f[i][j][] += f[i - ][j][]) %= mod;
            (f[i][j][] += f[i - ][j][]) %= mod;
            //for(int k = 0; k <= 3; k++)//��ǰ״̬ 

        }
    }
    int ans = (f[N][K][] + f[N][K][] % mod + f[N][K][] % mod + f[N][K][] % mod) % mod;
    cout << ans;
    return ;
}
/*
2 2 1
1 1
2 1 1
*/

D

F. The Shortest Statement

https://www.cnblogs.com/zwfymqz/p/9688315.html