A 不想走

求一下最短的距离就好了

signed main() {
    IOS;
    int a, b, x, y;
    cin >> a >> b >> x >> y;
    cout << min({abs(b - a), abs(a - x) + abs(y - b), abs(a - y) + abs(b - x)}) << endl;

    return 0;
}

B 连续数求和

直接暴力就好了

int main() {
	int T;
	cin>>T;
	for(int i=1;i<=100000;i++)
	{
		a[i]=a[i-1]+i;
	}
	for(int i=1;i<=T;i++)
	{
		cin>>n;
		for(int j=2;j<=100000;j++)
		{
			if(n<a[j])
			{
				cout<<"IMPOSSIBLE\n";
				break;
			}
			else
			{
				if(n%j==a[j]%j)
				{
					int tmp=n/j-a[j]/j;
					cout<<n<<" = ";
					for(int k=1;k<j;k++)
					{
						cout<<k+tmp<<" + ";
					}
					cout<<j+tmp<<"\n";
					break;
				}
			}
		}
	}
	return 0;
}

C 琴魔

用链表求一下所有的联通快就好了，注意是只有一个环还是多出来一块，要分类讨论一下

#define MAX 110
int a[MAX], in[MAX], e[MAX], st[MAX];
int n;

signed main() {
    IOS;
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    sort(a + 1, a + 1 + n);
    e[1] = 2, e[n] = n - 1, in[2]++, in[n - 1]++;
    for (int i = 2; i < n; i++) {
        if (a[i] - a[i - 1] <= a[i + 1] - a[i]) {
            e[i] = i - 1, in[i - 1]++;
        } else {
            e[i] = i + 1, in[i + 1]++;
        }
    }
    ll res = 0;
    for (int i = 1; i <= n; i++) {
        if (!st[i] && !in[i]) {
            int E = i;
            while (E) {
                if (st[E]) break;
                st[E] = 1;
                E = e[E];
            }
            res++;
        }
    }
     for (int i = 1; i <= n; i++) {
        if (!st[i]) {
            int E = i;
            while (E) {
                if (st[E]) break;
                st[E] = 1;
                E = e[E];
            }
            res++;
        }
    }
    cout << res << endl;
    return 0;
}

D. 素数分解

写个筛法跑一遍背包就好了

#define MAX 20005
int st[MAX], p[MAX], cnt;
int DP[MAX];
int n;
void Eratosthenes(){
    for(int i=2; i<=n;i++){
        if(!st[i]){
            p[++cnt]=i;
        }
        for(int j=1;j<=cnt&&i*p[j]<=n;j++){
            st[i*p[j]]=1;
            if(i%p[j]==0){
                break;
            }
        }
    }
}

signed main() {
    IOS;
    cin >> n;
    Eratosthenes();
    fill(DP, DP + n + 1, -1e9);
    DP[0] = 0;
    for (int i = 1; i <= cnt; i++) {
        int a = p[i];
        //cout << a << endl;
        for (int j = n; j >= a; j--) {
            DP[j] = max(DP[j], DP[j - a] + 1);
            //cout << DP[j] << ' ';
        }
        //cout << endl;
    }
    cout << DP[n] << endl;
    return 0;
}

E 勇者斗恶龙

重点是要先枚举位置再枚举装备，再更新完装备后再看看能扔几件装备

#define MAX 20005
int st[MAX], p[MAX], cnt;
int DP[MAX];
int n;
void Eratosthenes(){
    for(int i=2; i<=n;i++){
        if(!st[i]){
            p[++cnt]=i;
        }
        for(int j=1;j<=cnt&&i*p[j]<=n;j++){
            st[i*p[j]]=1;
            if(i%p[j]==0){
                break;
            }
        }
    }
}

signed main() {
    IOS;
    cin >> n;
    Eratosthenes();
    fill(DP, DP + n + 1, -1e9);
    DP[0] = 0;
    for (int i = 1; i <= cnt; i++) {
        int a = p[i];
        //cout << a << endl;
        for (int j = n; j >= a; j--) {
            DP[j] = max(DP[j], DP[j - a] + 1);
            //cout << DP[j] << ' ';
        }
        //cout << endl;
    }
    cout << DP[n] << endl;
    return 0;
}

F 公平正义

const int N = 20, V =3005;
int DP[N][V][2];
int num[N];
int NP;

int dfs(int n, int now, int lim) {
    if (now < 0) {
        return 0;
    }
    if(n == 0) return now == 0;
    if (~DP[n][now][lim]) return DP[n][now][lim];
    int res = 0,UP = lim ? num[n] : 9;
    for (int i = 0; i <= UP; i++) {
        res += dfs(n - 1, now + (n - NP) * i, lim && i == UP);
    }
    return DP[n][now][lim] = res;
}

int solve(int n) {
    int cnt = 0;
    while (n) {
        num[++cnt] = n % 10;
        n /= 10;
    }
    int res = 0;
    for (int i = 1; i <= cnt; i++) {
        memset(DP, -1, sizeof(DP));
        NP = i;
        res += dfs(cnt, 0, 1);
    }
    return res - cnt + 1;
}

signed main() {
    IOS;
    int L, R;
    cin >> L >> R;
    cout << solve(R) - solve(L - 1) << endl;
    return 0;
}