LevOJ - 第三期

LevOJ平台.sln

P2097 乘法口诀数列

问题描述

解决方法

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
#include <stdio.h>

int main() {
    int ans[100];
    int a1,a2,n;
    scanf("%d%d%d",&a1,&a2,&n);
    ans[0]=a1;
    ans[1]=a2;
    int j=2;
    for(int i=0;i<10;i++){
        if (ans[i]*ans[i+1]>9)
        {
            ans[j]=(ans[i]*ans[i+1])/10;
            j++;
            ans[j]=(ans[i]*ans[i+1])%10;
            j++;
        }
        else
        {
            ans[j]=ans[i]*ans[i+1];
            j++;
        }
    }
    for(int i=0;i<n;i++){
        printf("%d ",ans[i]);
    }
    printf("\n");
    return 0;
}

P2098 试试手气

问题描述

解决方法

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
#include <stdio.h>

int main()
{
    int index[6][6] = {0};
    int n;
    for (int i = 0; i < 6; i++)
    {
        scanf("%d", &index[0][i]);
    }
    scanf("%d", &n);
    for (int i = 0; i < 6; i++)
    {
        int temp = 6;
        for (int j = 1; j <= 5;)
        {
            if (temp==index[0][i])
            {
                temp--;
            }
            else
            {
                index[j][i] = temp;
                temp--;
                j++;
            }
        }
    }
    for (int i = 0; i < 6; i++)
    {
        printf("%d ", index[n][i]);
    }
    printf("\n");
    return 0;
}

P2099 机工士姆斯塔迪奥

问题描述

解决方法

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
#include <stdio.h>

int main() {
    int n, m, s, map[100][100]={0};
    scanf("%d%d%d", &n, &m, &s);
    int rc, ind, cnt=0;
    for(int i=0; i<s; i++) {
        scanf("%d%d", &rc, &ind);
        switch (rc)
        {
        case 0:
            for (int i = 0; i < m; i++)
            {
                map[ind-1][i] = 1;
            }
            break;
        case 1:
            for (int i = 0; i < n; i++)
            {
                map[i][ind-1] = 1;
            }
            break;

        default:
            break;
        }
    }
    for(int i=0; i<n; i++) {
        for(int j=0; j<m; j++) {
            if(map[i][j] == 0) {
                cnt++;
            }
        }
    }
    printf("%d", cnt);
    return 0;
}

P1992 猜帽子游戏

问题描述

输入输出样例

解决方法

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
#include <stdio.h>

int main()
{
    int n;
    scanf("%d", &n);
    int hats[100];
    for (int i = 0; i < n; i++)
        scanf("%d", &hats[i]);

    int k;
    scanf("%d", &k);
    for (int i = 0; i < k; i++)
    {
        int guesses[100];
        int ycnt = 0, ncnt = 0, hcnt = 0;

        for (int j = 0; j < n; j++)
        {
            scanf("%d", &guesses[j]);
            if (guesses[j] != 0)
            {                 
                hcnt = 1;
                if (guesses[j] == hats[j])
                {
                    ycnt++;
                }
                else
                {
                    ncnt++;
                }
            }
        }

        if (ncnt == 0 && ycnt > 0 && hcnt)
        {
            printf("Da Jiang!!!\n");
        }
        else
        {
            printf("Ai Ya\n");
        }
    }

    return 0;
}

P2023 超人与奥特曼

问题描述

解决方法

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
#include <stdio.h>

enum
{
    M = 100
};

int main()
{
    int x[M];
    int aotoaman = 0, surperman = 0;

    int n, k = 0;
    scanf("%d", &n);

    int l = n - 1;
    for (int i = 0; i < n; i++)
    {
        scanf("%d", &x[i]);
    }
    while (k != l + 1)
    {
        if (aotoaman >= surperman)
        {
            surperman += x[k++];
        }
        else if (aotoaman < surperman)
        {
            aotoaman += x[l--];
        }
    }
    printf("%d %d", surperman, aotoaman);

    return 0;
}

P1296 分形宇宙

问题描述

解决方法

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
#include <stdio.h>
#include <math.h>

char canvas[10000][10000];
void initCanvas(int size)
{
    for (int i = 0; i < size; i++)
    {
        for (int j = 0; j < size; j++)
        {
            canvas[i][j] = ' ';
        }
    }
}

void fra(int n, int x, int y)
{
    int size = pow(3, n - 2);
    if (n == 1)
    {
        canvas[x][y] = 'X';
        return;
    }
    fra(n - 1, x, y);
    fra(n - 1, x + 2 * size, y);
    fra(n - 1, x + size, y + size);
    fra(n - 1, x, y + 2 * size);
    fra(n - 1, x + 2 * size, y + 2 * size);
}

int main()
{
    int n;
    scanf("%d", &n);
    int size = pow(3, n - 1);
    initCanvas(size);
    fra(n, 0, 0);
    
    for (int i = 0; i < size; i++)
    {
        for (int j = 0; j < size; j++)
        {
            printf("%c", canvas[i][j]);
        }
        printf("\n");
    }

    return 0;
}

这是一道非常有趣的题 可玩性非常高~

有关分形，看看这篇介绍

有关本题，可以看看这篇思路讲解

另 （不建议使用的）偷鸡法

点击展开代码 
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
#include <stdio.h>

int main()
{
    int n;
    scanf("%d", &n);
    
    switch (n)
    {
    case 1:
        printf(
            "X\n");
        break;
    case 2:
        printf(
            "X X\n"
            " X \n"
            "X X\n");
        break;
    case 3:
        printf(
            "X X   X X\n"
            " X     X \n"
            "X X   X X\n"
            "   X X   \n"
            "    X    \n"
            "   X X   \n"
            "X X   X X\n"
            " X     X \n"
            "X X   X X\n");
        break;
    case 4:
        printf(
            "X X   X X         X X   X X\n"
            " X     X           X     X \n"
            "X X   X X         X X   X X\n"
            "   X X               X X   \n"
            "    X                 X    \n"
            "   X X               X X   \n"
            "X X   X X         X X   X X\n"
            " X     X           X     X \n"
            "X X   X X         X X   X X\n"
            "         X X   X X         \n"
            "          X     X          \n"
            "         X X   X X         \n"
            "            X X            \n"
            "             X             \n"
            "            X X            \n"
            "         X X   X X         \n"
            "          X     X          \n"
            "         X X   X X         \n"
            "X X   X X         X X   X X\n"
            " X     X           X     X \n"
            "X X   X X         X X   X X\n"
            "   X X               X X   \n"
            "    X                 X    \n"
            "   X X               X X   \n"
            "X X   X X         X X   X X\n"
            " X     X           X     X \n"
            "X X   X X         X X   X X\n");
        break;
    case 5:
        printf(
            "X X   X X         X X   X X                           X X   X X         X X   X X\n"
            " X     X           X     X                             X     X           X     X \n"
            "X X   X X         X X   X X                           X X   X X         X X   X X\n"
            "   X X               X X                                 X X               X X   \n"
            "    X                 X                                   X                 X    \n"
            "   X X               X X                                 X X               X X   \n"
            "X X   X X         X X   X X                           X X   X X         X X   X X\n"
            " X     X           X     X                             X     X           X     X \n"
            "X X   X X         X X   X X                           X X   X X         X X   X X\n"
            "         X X   X X                                             X X   X X         \n"
            "          X     X                                               X     X          \n"
            "         X X   X X                                             X X   X X         \n"
            "            X X                                                   X X            \n"
            "             X                                                     X             \n"
            "            X X                                                   X X            \n"
            "         X X   X X                                             X X   X X         \n"
            "          X     X                                               X     X          \n"
            "         X X   X X                                             X X   X X         \n"
            "X X   X X         X X   X X                           X X   X X         X X   X X\n"
            " X     X           X     X                             X     X           X     X \n"
            "X X   X X         X X   X X                           X X   X X         X X   X X\n"
            "   X X               X X                                 X X               X X   \n"
            "    X                 X                                   X                 X    \n"
            "   X X               X X                                 X X               X X   \n"
            "X X   X X         X X   X X                           X X   X X         X X   X X\n"
            " X     X           X     X                             X     X           X     X \n"
            "X X   X X         X X   X X                           X X   X X         X X   X X\n"
            "                           X X   X X         X X   X X                           \n"
            "                            X     X           X     X                            \n"
            "                           X X   X X         X X   X X                           \n"
            "                              X X               X X                              \n"
            "                               X                 X                               \n"
            "                              X X               X X                              \n"
            "                           X X   X X         X X   X X                           \n"
            "                            X     X           X     X                            \n"
            "                           X X   X X         X X   X X                           \n"
            "                                    X X   X X                                    \n"
            "                                     X     X                                     \n"
            "                                    X X   X X                                    \n"
            "                                       X X                                       \n"
            "                                        X                                        \n"
            "                                       X X                                       \n"
            "                                    X X   X X                                    \n"
            "                                     X     X                                     \n"
            "                                    X X   X X                                    \n"
            "                           X X   X X         X X   X X                           \n"
            "                            X     X           X     X                            \n"
            "                           X X   X X         X X   X X                           \n"
            "                              X X               X X                              \n"
            "                               X                 X                               \n"
            "                              X X               X X                              \n"
            "                           X X   X X         X X   X X                           \n"
            "                            X     X           X     X                            \n"
            "                           X X   X X         X X   X X                           \n"
            "X X   X X         X X   X X                           X X   X X         X X   X X\n"
            " X     X           X     X                             X     X           X     X \n"
            "X X   X X         X X   X X                           X X   X X         X X   X X\n"
            "   X X               X X                                 X X               X X   \n"
            "    X                 X                                   X                 X    \n"
            "   X X               X X                                 X X               X X   \n"
            "X X   X X         X X   X X                           X X   X X         X X   X X\n"
            " X     X           X     X                             X     X           X     X \n"
            "X X   X X         X X   X X                           X X   X X         X X   X X\n"
            "         X X   X X                                             X X   X X         \n"
            "          X     X                                               X     X          \n"
            "         X X   X X                                             X X   X X         \n"
            "            X X                                                   X X            \n"
            "             X                                                     X             \n"
            "            X X                                                   X X            \n"
            "         X X   X X                                             X X   X X         \n"
            "          X     X                                               X     X          \n"
            "         X X   X X                                             X X   X X         \n"
            "X X   X X         X X   X X                           X X   X X         X X   X X\n"
            " X     X           X     X                             X     X           X     X \n"
            "X X   X X         X X   X X                           X X   X X         X X   X X\n"
            "   X X               X X                                 X X               X X   \n"
            "    X                 X                                   X                 X    \n"
            "   X X               X X                                 X X               X X   \n"
            "X X   X X         X X   X X                           X X   X X         X X   X X\n"
            " X     X           X     X                             X     X           X     X \n"
            "X X   X X         X X   X X                           X X   X X         X X   X X\n");
        break;
    default:
        break;
    }
    return 0;
}

P2100 考试座位号

问题描述

解决方法

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
#include <stdio.h>

void printInfo(long long int info[][3], int m, int n)
{
    for (int i = 0; i < n; i++)
        if (m == (int)info[i][1])
            printf("%lld %lld\n", info[i][0], info[i][2]);
}

int main()
{
    int n;
    scanf("%d", &n);
    long long int info[10][3];
    for (int i = 0; i < n; i++)
        scanf("%lld %lld %lld", &info[i][0], &info[i][1], &info[i][2]);
    int m;
    scanf("%d", &m);
    int query[10];
    for (int i = 0; i < m; i++)
        scanf("%d", &query[i]);
    for (int i = 0; i < m; i++)
        printInfo(info, query[i], n);

    return 0;
}

P2101 一帮一

问题描述

解决方法

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
#include <stdio.h>

int main()
{
    int n;
    scanf("%d", &n);
    int gender[50];
    char name[50][20];
    int isUsed[50] = {0};
    for (int i = 0; i < n; i++)
    {
        scanf("%d", &gender[i]);
        scanf("%s", name[i]);
    }

    for (int i = 0; i < n; i++)
    {
        if (isUsed[i])
        {
            continue;
        }

        for (int j = n - 1; j >= 0; j--)
        {
            if (i != j && !isUsed[j] && gender[i] != gender[j])
            {
                printf("%s %s\n", name[i], name[j]);
                isUsed[i] = 1;
                isUsed[j] = 1;
                break;
            }
        }
    }

    return 0;
}

P1325 区间合并

问题描述

解决方法

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
#include <stdio.h>

void sort(int s[], int e[], int n)
{
    for (int i = 0; i < n - 1; i++)
        for (int j = 0; j < n - 1 - i; j++)
            if (s[j] > s[j + 1])
            {
                int temps = s[j];
                int tempe = e[j];
                s[j] = s[j + 1];
                e[j] = e[j + 1];
                s[j + 1] = temps;
                e[j + 1] = tempe;
            }
}

void merge(int s[], int e[], int n)
{
    int start = s[0], end = e[0];
    for (int i = 1; i < n; i++)
    {
        if (end < s[i])
        {
            printf("%d %d\n", start, end);
            start = s[i];
            end = e[i];
        }
        else if (end < e[i])
        {
            end = e[i];
        }
    }
    printf("%d %d\n", start, end);
}

int main()
{
    int n;
    scanf("%d", &n);
    int s[n], e[n];
    for (int i = 0; i < n; i++) scanf("%d %d", &s[i], &e[i]);
    sort(s, e, n);
    merge(s, e, n);

    return 0;
}

P2045 数字蚯蚓

问题描述

输入输出样例

解决方法

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
#include <stdio.h>
#include <string.h>

int main()
{
    char str[200], be[200], af[200];
    fgets(str, 200, stdin);
    int len = strlen(str);
    int k;
    scanf("%d", &k);

    strcpy(af, str);
    for (int i = 0; i < k; i++)
    {
        af[i] = af[k];
    }
    strcpy(be, str);
    for (int i = k; i < len - 1; i++)
    {
        be[i] = be[k - 1];
    }
    printf("%s", be);
    printf("%s", af);
    return 0;
}

第三期完结

看完了点个赞再走嘛




__EOF__\_\_EOF\_\_


OJ题解
#LevOJ #C语言
LevOJ - 第三期
https://churk.top/2024/11/10/2024-11-10-01/
作者
Churk Ben
发布于
2024年11月10日
许可协议