A、数字计数

直接排序，注意要去掉重复，样例坑了一手。
其他根据题意翻译就行了

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define ms(__x__,__val__) memset(__x__, __val__, sizeof(__x__))
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x, int op = 10) { if (!x) { putchar('0'); if (op)    putchar(op); return; }    char F[40]; ll tmp = x > 0 ? x : -x;    if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]);    if (op)    putchar(op); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }
const int dir[][2] = { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;

const int N = 100 + 7;
vector<int> a;

int main() {
    int n = read();
    for (int i = 1; i <= n; ++i) {
        a.push_back(read());
    }
    sort(all(a));
    a.erase(unique(all(a)), a.end());
    n = a.size();
    print(a[n - 1] - a[n - 2], 32);
    print(a[n - 1] - a[1], 32);
    print(a[n - 2] - a[1], 32);
    print(a[n - 2] - a[0], 32);
    return 0;
}

#pragma GCC target("avx,sse2,sse3,sse4,popcnt") #pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math") #include <bits/stdc++.h> using namespace std; #define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0) #define all(vv) (vv).begin(), (vv).end() #define endl "\n" #define pai pair<int, int> #define ms(x,val) memset(x, val, sizeof(x)) typedef long long ll; typedef unsigned long long ull; typedef long double ld; inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar()) s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; } inline void print(ll x, int op = 10) { if (!x) { putchar('0'); if (op) putchar(op); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-'); int cnt = 0; while (tmp > 0) { F[cnt++] = tmp % 10 + '0'; tmp /= 10; } while (cnt > 0)putchar(F[--cnt]); if (op) putchar(op); } inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; } ll qpow(ll a, ll b) { ll ans = 1; while (b) { if (b & 1) ans = a; b >>= 1; a = a; } return ans; } ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans = a) %= mod; b >>= 1; (a = a) %= mod; }return ans % mod; } inline int lowbit(int x) { return x & (-x); } const int dir[][2] = { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} }; const int MOD = 1e9 + 7; const int INF = 0x3f3f3f3f; const int N = 1000 + 7; int a[N], vis[N]; int main() { int n = read(); for (int i = 1; i <= n; ++i) a[i] = read(); int ans = 0; for (int i = 1; i <= n; ++i) { int sum = 0; ms(vis, 0); for (int j = i; j <= n; ++j) { if (vis[a[j]] == 0) ++sum; ++vis[a[j]]; ans += sum; } } print(ans); return 0; }

牛客IOI周赛18-普及组

A、数字计数