A. Diverse Substring

題意：找一個子串滿足任何一個字元的出現次數小於 n/2 n是子串的長度，

顯然兩個不相同的字元組成的子串就是滿足情況的；

#include<bits/stdc++.h>

using namespace std;

#define out fflush(stdout)
#define fast ios::sync_with_stdio(0),cin.tie(0);

#define FI first
#define SE second

typedef long long ll;
typedef pair<int,int> P;

const int maxn = 1000 + 7;
const int INF = 0x3f3f3f3f;


int n;
char s[maxn];
map<char, int> mp;

int main() {
    scanf("%d", &n);
    scanf("%s", s+1);

    for(int i = 2; i <= n; ++i) {
        if(s[i] != s[i-1]) {
            puts("YES");
            printf("%c%c", s[i-1], s[i]);
            return 0;
        }
    }
    puts("NO");
    return 0;
}
 

B. Vasya and Books

題意：給定n本書的位置，從上往下，編號1~n不重複，沒找一本書需要把他上面的書全搬走，問每找一本書需要搬幾本書，如果之前已經搬出來的書不用搬動，輸出0；

思路：直接模擬

#include<bits/stdc++.h>

using namespace std;

#define out fflush(stdout)
#define fast ios::sync_with_stdio(0),cin.tie(0);

#define FI first
#define SE second

typedef long long ll;
typedef pair<int,int> P;

const int maxn = 2e5 + 7;
const int INF = 0x3f3f3f3f;


int n;
int a[maxn], b[maxn];

int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i) {
        scanf("%d", &a[i]);
    }
    for(int i = 1; i <= n; ++i) {
        scanf("%d", &b[i]);
    }
    int id = 1;
    set<int> st;
    for(int i = 1; i <= n; ++i) {
        if(st.count(b[i])) {
            printf("%d%c", 0, (i == n ? '\n' : ' '));
        }
        else {
            int ans = 1;
            while(a[id] != b[i]) {
                st.insert(a[id]);
                id++, ans++;
            }
            id++;
            printf("%d%c", ans, (i == n ? '\n' : ' '));
        }
    }


    return 0;
}
 

C. Vasya and Robot

題解：https://blog.csdn.net/xiang_6/article/details/83686299

D. Berland Fair

題解：https://blog.csdn.net/xiang_6/article/details/83686379

E. Segment Sum

題解：