Education CodeForces Round 63 Div.2

A. Reverse a Substring

代码：

#include <bits/stdc++.h>
using namespace std;

int N;
string s;

int main() {
    scanf("%d", &N);
    cin >> s;

    int temp = -1;
    for(int i = 1; s[i]; i ++) {
        if(s[i] < s[i - 1]) {
            temp = i;
            break;
        }
    }

    if(temp == -1) printf("NO\n");
    else {
        printf("YES\n");
        temp += 1;
        printf("%d %d\n", temp - 1, temp);
    }

    return 0;
}

B. Game with Telephone Numbers

代码：

#include <bits/stdc++.h>
using namespace std;

int N;
int eight = 0;
string s;

int main() {
    scanf("%d", &N);
    cin >> s;
    bool flag = true;
    for(int i = 0; i <= N - 11; i ++) {
        if(s[i] == '8')
            eight ++;
    }
    if(eight <= ((N - 11) / 2)) flag = false;

    if(flag) printf("YES\n");
    else printf("NO\n");

    return 0;
}

C. Alarm Clocks Everywhere

代码：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 3e5 + 10;
int N, M;
long long a[maxn], p[maxn], b[maxn];

long long gcd(long long a, long long b) {
    return b == 0 ? a : gcd(b, a % b);
}

int main() {
    scanf("%d%d", &N, &M);
    long long t;
    for(int i = 1; i <= N; i ++) {
        cin >> a[i];
        //scanf("%lld", &a[i]);
        if(i == 1) b[i] = 0;
        else b[i] = a[i] - a[i - 1];
    }

    for(int i = 2; i <= N; i ++) {
        if(i == 2) t = b[i];
        else t = gcd(t, b[i]);
    }

    bool flag = false;
    int temp;
    for(int i = 1; i <= M; i ++) {
        cin >> p[i];
        //scanf("%lld", &p[i]);
        if(t % p[i] == 0) {
            temp = i;
            flag = true;
        }
    }

    int ansp = temp;
    if(!flag) printf("NO\n");
    else {
        printf("YES\n");
        cout << a[1] << " " << ansp << endl;
    }

    return 0;
}

D. Beautiful Array

代码：（D 神仙 dp 自己写了贪心 但是不对有一些情况不可以 枯了 ）

#include <bits/stdc++.h>
using namespace std;

const int maxn = 3e5 + 10;
long long a[maxn], dp[maxn][5];
int N, K;

int main() {
    memset(dp, -1e18, sizeof(dp));
    scanf("%d%d", &N, &K);
    long long ans = 0;
    for(int i = 1; i <= N; i ++) {
        cin >> a[i];
        dp[i][0] = max(dp[i - 1][0], 0LL) + a[i];
        dp[i][1] = max(max(dp[i - 1][1], dp[i - 1][0]), 0LL) + 1LL * K * a[i];
        dp[i][2] = max(max(max(dp[i - 1][1], dp[i - 1][0]), dp[i - 1][2]), 0LL) + a[i];
        ans = max(ans, max(dp[i][0], max(dp[i][1], dp[i][2])));
    }
    cout << ans << endl;
    return 0;
}