A.游游的整数翻转

用 string 存，模拟即可。

Code:

#include <bits/stdc++.h>
using namespace std;
int main() {
    string s;
    int n; bool ok = 0;
    cin >> s >> n;
    for (int i = n - 1; i >= 0; i -- ) {
        if (s[i] == '0' && ok == 0);
        else cout << s[i], ok = 1;
    }
    for (int i = n; i < s.size(); i ++ ) cout << s[i];
}

B.游游的排列统计

很小，，计算量最多为 ，况且其中有回溯现象，所以计算量大约不超过 。

Code:

#include <bits/stdc++.h>
#define int long long
using namespace std;
int n, ans, a[1001], d[1001];
bool check(int x) {
    if (x == 1) return true;
    for (int i = 2; i * i <= x; i ++ )
        if (x % i == 0)
            return true;
    return false;
}
void dfs(int x) {
    if (x > n) {
        ans ++ ;
        return;
    }
    for (int i = 1; i <= n; i ++ )
        if (!d[i]) {
            d[i] = 1;
            if (x == 1 || check(i + a[x - 1])) {
                a[x] = i;
                dfs(x + 1);
            }
            d[i] = 0;
        }
}
signed main() {
    cin >> n;
    dfs(1);
    cout << ans;
}

C.游游刷题

模拟。

每次对 模 ，若为 ，答案直接加一；否则加入待定数组 中。

若有一天做 和 合法，则保证 + ，删去 ，答案加一。

Code:

#pragma GCC optimize(2)
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 1e5 + 10;
int n, k, a[N], ans;
int e[N]; int l;
signed main() {
    cin >> n >> k;
    for (int i = 1; i <= n; i ++ ) {
        cin >> a[i];
        a[i] %= k;
        if (a[i] == 0) ans ++ ;
        else e[ ++ l] = a[i];
    }
    for (int i = 1; i <= l; i ++ )
        for (int j = i + 1; j <= l; j ++ )
            if (e[i] + e[j] == k) {
                ans ++ ;
                swap(e[j], e[l]);
                l -- ;
                swap(e[i], e[l]);
                l -- ;
                i -- ;
                break;
            }
    cout << ans;
}

D.游游买商品

背包dp即可

Code:

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 1e3 + 10;
int n, x, f[N][N][3], b[N], a[N]; 
signed main() {
    scanf("%lld%lld", &n, &x);
    for (int i = 1; i <= n; i ++ )
        scanf("%lld", &a[i]);
    for (int i = 1; i <= n; i ++ )
        scanf("%lld", &b[i]);
    memset(f, -0x3f, sizeof f);
    f[0][0][0] = 0;
    int res = 0;
    for (int i = 1; i <= n; i ++ ) {
        for (int j = 0; j <= x; j ++ ) {
            f[i][j][0] = max(f[i - 1][j][0], max(f[i - 1][j][1], f[i - 1][j][2]));
            if (j >= a[i] / 2)
                f[i][j][1] = max(f[i][j][1], f[i - 1][j - a[i] / 2][2] + b[i]);
            if (j >= a[i])
                f[i][j][2] = max(f[i - 1][j - a[i]][0], max(f[i - 1][j - a[i]][1], f[i - 1][j - a[i]][2])) + b[i];
            res = max(res, max(f[i][j][0], max(f[i][j][1], f[i][j][2])));
        }
    }
    cout << res;
}