At grand 024

A

/* Huyyt */
#include <bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
#define mkp(a,b) make_pair(a,b)
#define pb push_back
const int dir[][] = {{, }, {, }, {, -}, { -, }, {, }, {, -}, { -, -}, { -, }};
using namespace std;
typedef long long ll;
inline void read(int &v)
{
        v = ;
        char c = ;
        int p = ;
        while (c < '' || c > '')
        {
                if (c == '-')
                {
                        p = -;
                }
                c = getchar();
        }
        while (c >= '' && c <= '')
        {
                v = (v << ) + (v << ) + c - '';
                c = getchar();
        }
        v *= p;
}
const long long mod = 1e9 + ;
const int N = 1e5 + ;
int main()
{
        ll a, b, c, k;
        cin >> a >> b >> c >> k;
        if (k %  == )
        {
                cout << b - a << endl;
        }
        else
        {
                cout << a - b << endl;
        }
        return ;
}

B

只要算出最长 数字连续子序列就可以了

/* Huyyt */
#include <bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
#define mkp(a,b) make_pair(a,b)
#define pb push_back
const int dir[][] = {{, }, {, }, {, -}, { -, }, {, }, {, -}, { -, -}, { -, }};
using namespace std;
typedef long long ll;
inline void read(int &v)
{
        v = ;
        char c = ;
        int p = ;
        while (c < '' || c > '')
        {
                if (c == '-')
                {
                        p = -;
                }
                c = getchar();
        }
        while (c >= '' && c <= '')
        {
                v = (v << ) + (v << ) + c - '';
                c = getchar();
        }
        v *= p;
}
const long long mod = 1e9 + ;
const int N = 2e5 + ;
int n;
int num[N];
int aim[N];
int main()
{
        read(n);
        int now;
        for (int i = ; i <= n; i++)
        {
                read(now);
                num[now] = i;
        }
        int ans = ;
        int ansmaxn = ;
        for (int i = ; i <= n; i++)
        {
                if (num[i] > num[i - ])
                {
                        ans++;
                }
                else
                {
                        ans = ;
                }
                ansmaxn = max(ansmaxn, ans);
        }
        cout << n - ansmaxn << endl;
        return ;
}

C

给你两个数组A,B  A全为0  B是给定的N个数

一次操作可以使得A_i(2<=i<=N) 变为 A_i-1+1

解:

首先排除不合法的情况: ①.当A₀不为0的时候 ②.当前面的数减后面的数大于1的时候

/* Huyyt */
#include <bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
#define mkp(a,b) make_pair(a,b)
#define pb push_back
const int dir[][] = {{, }, {, }, {, -}, { -, }, {, }, {, -}, { -, -}, { -, }};
using namespace std;
typedef long long ll;
inline void read(int &v)
{
        v = ;
        char c = ;
        int p = ;
        while (c < '' || c > '')
        {
                if (c == '-')
                {
                        p = -;
                }
                c = getchar();
        }
        while (c >= '' && c <= '')
        {
                v = (v << ) + (v << ) + c - '';
                c = getchar();
        }
        v *= p;
}
const long long mod = 1e9 + ;
const int N = 2e5 + ;
int n;
ll num[N];
int main()
{
        read(n);
        for (int i = ; i <= n; i++)
        {
                cin >> num[i];
        }
        if (num[] != )
        {
                cout << - << endl;
                return ;
        }
        for (int i = ; i <= n; i++)
        {
                if (num[i] - num[i - ] > )
                {
                        cout << - << endl;
                        return ;
                }
        }
        ll anser = ;
        for (int i = ; i <= n; i++)
        {
                if (num[i] == num[i - ] + )
                {
                        anser++;
                }
                else
                {
                        anser += num[i];
                }
        }
        cout << anser << endl;
        return ;
}

D