Codeforces Edu Round 63 A-E

A. Reverse a Substring

容易看出，只要符合递增顺序就符合\(NO\)，否则则可以找到一组，每次记录最大值比较即可。

#include <cstdio>
#include <iostream>
using namespace std;
const int N = 300010;
int n;
char s[N];
int main(){
    scanf("%d%s", &n, s + 1);
    int maxn = s[1], k = 1;
    for(int i = 2; i <= n; i++){
        if(s[i] < maxn){
            printf("YES\n%d %d\n", k, i);
            return 0;
        }
        if(s[i] > maxn){
            maxn = s[i];
            k = i;
        }
    }
    printf("NO\n");
    return 0;
} 

B. Game with Telephone Numbers

\(Vasya\)用尽所有轮次可以把几个\(8\)扔到最前面，只要这个数大于轮次，就说明\(Petya\)无法逆天改命。

#include <cstdio>
#include <iostream>
#include <cstring> 
using namespace std;
const int N = 100010;
int n;
char s[N];
int main(){
    scanf("%d%s", &n, s + 1);
    int round = (n - 11) >> 1;
    int cnt = 0; 
    for(int i = 1; i <= round && round <= n; i++){
        if(s[i] == '8')cnt++, round ++;
    }
    int i = round + 1;
    while(s[i] == '8' && i <= n) i++, cnt++;
    if(cnt > (n - 11) >> 1) puts("YES");
    else puts("NO");
    return 0;
}

C. Alarm Clocks Everywhere

显然，y设置为\(x[1]\)即可，因为即使设置在之前，还是要经过\(x[1]\)，还是以\(p\)往上跳，是没有区别的。

我们只要找到一个\(p_j\)，使其满足\(p_j | a[i + 1] - a[i] (1 <= i < n)\)。（整除的意思）

#include <cstdio>
#include <iostream>
using namespace std;
typedef long long LL;
const int N = 300010;
int n, m;
LL x[N], p[N];
LL gcd(LL a, LL b){
    return b ? gcd(b, a % b) : a;
}
int main(){
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) 
        scanf("%lld", x + i);
    for(int i = 1; i <= m; i++)
        scanf("%lld", p + i);
        
    LL ans = x[2] - x[1];
    for(int i = 2; i < n; i++)
        ans = gcd(ans, x[i + 1] - x[i]);
    
    for(int i = 1; i <= m; i++){
        if(ans % p[i] == 0){
            printf("YES\n%lld %d\n", x[1], i);
            return 0;
        }
    }
    printf("NO");
    return 0;
}

D. Beautiful Array

想到了对数列总和最大贡献，但是其实最大贡献不一定能最大化答案，于是惨遭\(WA10\)。

\(WA\)炸代码：

#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;
const int N = 300010, INF = 2147483647;
typedef long long LL;
int n, x, L[N];
LL a[N], f[N];
int main(){
    scanf("%d%d", &n, &x);
    for(int i = 1; i <= n; i++) 
        scanf("%lld", a + i);
    
    if(x < 0){
        f[1] = a[1];
        L[1] = 1;
        LL minn = f[1];
        int k = 1;
        for(int i = 2; i <= n; i++) {
            if(f[i - 1] + a[i] < a[i]){
                L[i] = L[i - 1];
                f[i] = f[i - 1] + a[i];
            }else{
                L[i] = i;
                f[i] = a[i];
            }
            
            if(f[i] < minn){
                minn = f[i], k = i;
            }
        }
        if(minn < 0){
            for(int i = L[k]; i <= k; i++) 
                a[i] *= x;
        } 
    }else{
        
        f[1] = a[1];
        L[1] = 1;
        LL maxn = f[1];
        int k = 1;
        for(int i = 2; i <= n; i++) {
            if(f[i - 1] + a[i] > a[i]){
                L[i] = L[i - 1];
                f[i] = f[i - 1] + a[i];
            }else{
                L[i] = i;
                f[i] = a[i];
            }
            
            if(f[i] > maxn){
                maxn = f[i], k = i;
            }
        }
        if(maxn > 0){
            for(int i = L[k]; i <= k; i++) 
                a[i] *= x;
        } 
        
    }
    
    f[1] = a[1];
    LL maxn = f[1];
    for(int i = 2; i <= n; i++){
        f[i] = max(f[i - 1] + a[i], a[i]);
        maxn = max(maxn, f[i]);
    }
    printf("%lld\n", max(maxn, 0ll));
    return 0;
}

看了题解直接自闭掉，实际上，可以设

\(f[0]\)为当前以\(i\)为右端点，未加入\(x\)的极大值，那么它必须大于\(0\)

\(f[1]\)代表开始连续计算\(* x\)了，如果这时候还没有\(f[0]\)大，那么它也没用了

\(f[2]\)代表乘完了的，探究是否还能再加入\(a[i]\)

如果\(x > 0\)，那么\(f[2] + a[i]\)一定不会最优

如果\(x < 0\)，有可能有这种情况，前面负数后面正数

#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;
const int N = 300010, INF = 2147483647;
typedef long long LL;
int n, x, L[N];
LL a[N], f[3], ans = 0;
int main(){
    scanf("%d%d", &n, &x);
    for(int i = 1; i <= n; i++) 
        scanf("%lld", a + i);
    
    for(int i = 1; i <= n; i++){
        f[0] = max(f[0] + a[i], 0ll);
        f[1] = max(f[0], f[1] + a[i] * x);
        f[2] = max(f[1], f[2] + a[i]);
        ans = max(ans, f[2]); 
    } 
    printf("%lld\n", ans);
    return 0;
}

E. Guess the Root

高斯消元\(/\)拉格朗日插值法都不会，咕咕咕。