2022 洛谷 SCP-J 第一轮模拟题（入门级 C++）

 1 | #include <iostream>
 2 | using namespace std;
 3 | const int MAXN = 1000050;
 4 | int n, a[MAXN], a1[MAXN], b[MAXN], lim;
 5 | void solve1() {
 6 |     for (int i = 1; i <= n; i++)
 7 |         b[a[i]]++; //①
 8 |     for (int i = 1; i <= lim; i++) {
 9 |         if (b[i]) //②
10 |             cout << i << " ";
11 |     }
12 |     cout << endl;
13 | }
14 | void solve2() {
15 |     int cnt = 0, flag;
16 |     for (int i = 1; i <= n; i++) {
17 |         flag = false;
18 |         for (int j = 1; j <= n - 1; j++) {
19 |             if (a[j] > a[j + 1]) {
20 |                 swap(a[j], a[j + 1]);
21 |                 cnt++;
22 |                 flag = true;
23 |             }
24 |         }
25 |         //if (flag==false)
26 |         // break;
27 |     }
28 |     for (int i = 1; i <= n; i++)
29 |         cout << a[i] << " ";
30 |     cout << endl;
31 | }
32 | int main() {
33 |     cin >> n;
34 |     for (int i = 1; i <= n; i++) {
35 |         cin >> a[i];
36 |         a1[i] = a[i]; //③
37 |         lim = max(a[i], lim); //④
38 |     }
39 |     solve1();
40 |     for (int i = 1; i <= n; i++)
41 |         a[i] = a1[i];
42 |     solve2();
43 |     return 0;
44 | }

已知 $1\le n,a_i\le10^6$。完成下列题目。

 1 | #include <cstdio>
 2 | #include <cstring>
 3 |
 4 | const int maxn = 1003;
 5 |
 6 | int type, n, m;
 7 | char s[maxn], t[maxn];
 8 |
 9 | int main() {
10 |     scanf("%d %s %s", &type, t, s);
11 |     n = strlen(s); m = strlen(t);
12 |     if (type < 2) {
13 |         for (int i = 0; i < m; ++i) s[i] = t[i];
14 |     } else if (type == 2) {
15 |         strcpy(s, t);
16 |         // 提示：如果此时调用 printf("%s\n", s)，则本次输出结果为整个 t 字符串和换行，没有其他字符。
17 |     } else {
18 |         for (int i = 0; i < m; i += 4) {
19 |             unsigned int code = 0, pos = i;
20 |             for (int j = 1; pos < i+4; j*=100, ++pos) {
21 |                 if (pos == m) break;
22 |                 code += t[pos] * j;
23 |             }
24 |             pos = i;
25 |             while (code != 0) {
26 |                 s[pos++] = code % 100;
27 |                 code /= 100;
28 |             }
29 |         }
30 |     }
31 |     for (int i = 0; i < n; ++i) printf("%c", s[i]);
32 |     printf("\n");
33 | }

输入保证 `t` 不长于 `s`，两字符串仅含大小写字母且非空，`type=1,2,3`。

 1 | #include <iostream>
 2 | using namespace std;
 3 |
 4 | const int INF = 1000000000;
 5 | #define Front 0
 6 | #define Back 1
 7 | #define Left 2
 8 | #define Right 3
 9 | #define Up 4
10 | #define Down 5
11 | int w[6], a[1003][1003];
12 | const int way1[] = {Up, Right, Down, Left};
13 | const int way2[] = {Up, Front, Down, Back};
14 | const int way3[] = {Left, Front, Right, Back};
15 | int get_max(int &a, int b) {
16 |     return a = max(a, b);
17 | }
18 | int right_rotate(int &u) {
19 |     for (int i = 0; i < 4; ++ i)
20 |         if (u == way1[i])
21 |             return u = way1[(i + 1) % 4];
22 |     return u;
23 | }
24 | int front_rotate(int &u) {
25 |     for (int i = 0; i < 4; ++ i)
26 |         if (u == way2[i])
27 |             return u = way2[(i + 1) % 4];
28 |     return u;
29 | }
30 | const int anchorX = Up;
31 | const int anchorY = Front;
32 | const int anchorZ = Right;
33 | int find_down(int u, int v) {
34 |     if (u == Down || u == Up) return anchorX ^ (u == Up);
35 |     if (v == Down || v == Up) return anchorY ^ (v == Up);
36 |     for (int i = 0; i < 4; ++ i)
37 |         if (u == way3[i])
38 |             return anchorZ ^ (v == way3[(i + 1) % 4]);
39 |     return -1;
40 | }
41 | int n, m, dp[1003][1003][6][6];
42 | int main() {
43 |     cin >> n >> m;
44 |     for (int i = 0; i < n; ++ i)
45 |         for (int j = 0; j < m; ++ j)
46 |             cin >> a[i][j];
47 |     for (int i = 0; i < 6; ++ i)
48 |         cin >> w[i];
49 |     for (int i = 0; i < n; ++ i)
50 |         for (int j = 0; j < m; ++ j)
51 |             for (int a = 0; a < 6; ++ a)
52 |                 for (int b = 0; b < 6; ++ b)
53 |                     dp[i][j][a][b] = -INF;
54 |     dp[0][0][anchorX][anchorY] = a[0][0] * w[Down];
55 |     for (int i = 0; i < n; ++ i)
56 |         for (int j = 0; j < m; ++ j)
57 |             for (int p = 0; p < 6; ++ p)
58 |                 for (int q = 0; q < 6; ++ q)
59 |                     if (dp[i][j][p][q] != -INF) {
60 |                         int x = dp[i][j][p][q];
61 |                         int u1 = p, v1 = q;
62 |                         right_rotate(u1);
63 |                         right_rotate(v1);
64 |                         get_max(dp[i][j + 1][u1][v1],
65 |                             x + w[find_down(u1, v1)] * a[i][j + 1]);
66 |                         int u2 = p, v2 = q;
67 |                         front_rotate(u2);
68 |                         front_rotate(v2);
69 |                         get_max(dp[i + 1][j][u2][v2],
70 |                             x + w[find_down(u2, v2)] * a[i + 1][j]);
71 |                     }
72 |     int ans = -INF;
73 |     for (int p = 0; p < 6; ++ p)
74 |         for (int q = 0; q < 6; ++ q)
75 |             ans = max(ans, dp[n - 1][m - 1][p][q]);
76 |     printf("%d\n", ans);
77 |     return 0;
78 | }

输入数据的绝对值均不超过 $10^3$。

有 $n$ 种各一张、面值为 $a_i$ 的纸币，求能支付多少种非零金额。$n\le200$，面值总和不超过 5000。

#include <iostream>
using namespace std;
const int MAXN = 210, MAXM = 5010;
int n, m, f[MAXM], a[MAXN];
int main() {
    cin >> n;
    for (int i = 1; i <= n; i++) { cin >> a[i]; ①; }
    ②;
    for (int i = 1; i <= n; i++)
        ③
            f[j] = ④;
    int ans = 0;
    for (int i = 1; i <= m; i++) if (⑤) ans++;
    cout << ans;
    return 0;
}

给 $n(1\le n\le20)$ 个互不相同的正整数 $a_i(1\le a_i\le10^9)$，在每个数前放 `+` 或 `-`，判断是否能使算式值为 17。

#include <cstdio>
using namespace std;
const int maxn = 25, aim = 17;
int n, a[maxn];
bool ans;
int getBit(const int s, int p) { return ①; }
int main() {
    scanf("%d", &n);
    for (②) scanf("%d", a + i);
    for (int s=0, upperBound=③; s<=upperBound; ++s) {
        ④;
        for (int j=0; j<n; ++j) if (getBit(s,j)==1) sum += a[j];
        else { ⑤; }
        if (int(sum)==aim) { ans=true; break; }
    }
    printf("%s\n", ans ? "Yes" : "No");
}