新生第二次考核代码

录播链接

A.HFUU(easy version)

字符串

#include <stdio.h>
char a[100];
int n = 9, m = 32;
int main()
{
    for (int i = 0; i < n; i++)
    {
        scanf("%s", a + 1); //字符串的读入是不需要加&符号的
        //用%s读字符串会读到空格或者换行符为止
        puts(a + 1); // puts可以输出一个字符串并且自带换行
    }
    return 0;
}

字符数组

#include <stdio.h>
char a[100];
int n = 9, m = 32;
int main()
{
    for (int i = 0; i < n; i++)
    {
        scanf("%s", a); //字符串的读入是不需要加&符号的
        puts(a);
    }
    return 0;
}

B.HFUU(hard version)

#include <stdio.h>
#include <string.h>

char a[110];
int n;

int main()
{
    scanf("%s", a); //让字符串从下标为1的地方开始
    n = strlen(a);  // string.h中的函数，用于获取字符串的长度
    for (int i = 0; i < n; i++)
    {
        printf("%c", a[i]);
        if (i < 6)
            continue; //结束当前轮循环，进入下一轮循环
        // ILovehfuuacm

        if (a[i - 6] == 'h' && a[i - 5] == 'f' && a[i - 4] == 'u' &&
            a[i - 3] == 'u' && a[i - 2] == 'a' && a[i - 1] == 'c' && a[i] == 'm')
        {
            printf("yyds");
        }
    }

    return 0;
}

C.HFUU(very hard version)

#include <stdio.h>
char a[1000][1000];
int n;
int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &n);
        for (int i = 1; i <= 13 * n + 19; i++)
        {
            a[1][i] = a[4 * n + 5][i] = '*';
        }
        for (int i = 1; i <= 4 * n + 5; i++)
        {
            a[i][1] = a[i][13 * n + 19] = '*';
        }
        for (int i = n + 2; i <= 3 * n + 4; i++)
        {
            a[i][n + 3] = a[i][3 * n + 5] = a[i][4 * n + 7] = a[i][7 * n + 11] = a[i][9 * n + 13] = a[i][10 * n + 15] = a[i][12 * n + 17] = '@';
        }
        for (int i = n + 3; i <= 3 * n + 5; i++)
        {
            a[2 * n + 3][i] = '@';
        }
        for (int i = 4 * n + 7; i <= 6 * n + 9; i++)
        {
            a[2 * n + 3][i] = a[n + 2][i] = '@';
        }
        for (int i = 7 * n + 11; i <= 9 * n + 13; i++)
        {
            a[3 * n + 4][i] = '@';
        }
        for (int i = 10 * n + 15; i <= 12 * n + 17; i++)
        {
            a[3 * n + 4][i] = '@';
        }
        for (int i = 1; i <= 4 * n + 5; i++)
        {
            for (int j = 1; j <= 13 * n + 19; j++)
            {
                if (a[i][j] != '@' && a[i][j] != '*')
                    a[i][j] = '.';
            }
        }
        for (int i = 1; i <= 4 * n + 5; i++)
        {
            for (int j = 1; j <= 13 * n + 19; j++)
            {
                printf("%c", a[i][j]);
                a[i][j] = '.';
            }
            puts("");
        }
    }
    return 0;
}

D.浅浅的解个方程

#include <stdio.h>
#define N 20001
#define ll long long
int t,n;
char s[N];

int main()
{
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d %s",&n,s);
        ll sig = 1,num = 0;
        ll liftSum = 0,rightSum = 0;
        ll sig_x = 1;
        int i;
        for (i=0;i<n;++i)
        {
            if (s[i]=='=')
            {
                liftSum += sig * num;
                break;
            }
            if (s[i]=='+'||s[i]=='-') liftSum += sig * num;
            if (s[i]=='-') sig = -1,num = 0;
            else if (s[i]=='+') sig = 1,num = 0;
            else if (s[i]=='x') sig_x = sig;
            else num *= 10,num += s[i] - '0';
        }
        sig = 1,num = 0;
        for (i=i+1;i<n;++i)
        {
            if (s[i]=='+'||s[i]=='-') rightSum += sig * num;
            if (s[i]=='-') sig = -1,num = 0;
            else if (s[i]=='+') sig = 1,num = 0;
            else if (s[i]=='x') sig_x = - sig;
            else num *= 10,num += s[i] - '0';
            if (i==n-1) rightSum += sig * num;
        }
        //printf("%lld",liftSum);
        printf("x=%lld\n",sig_x*(rightSum-liftSum));
    }
    return 0;
}

E.精打细算的小葛（easy version）

#include <stdio.h>
#include <stdlib.h>
int main()
{
    int n, a, b, p, q, ans = 0;
    scanf("%d%d%d%d%d", &n, &a, &b, &p, &q);
    for (int i = 1; i <= n; i++)
    {
        if (i % a == 0 && i % b == 0)
        {
            if (p > q)
                ans += p;
            else
                ans += q;
            continue;
        }
        if (i % a == 0)
            ans += p;
        if (i % b == 0)
            ans += q;
    }
    printf("%d", ans);
    return 0;
}

F.精打细算的小葛（hard version）

#include <stdio.h>
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        long long n, a, b, p, q, ans = 0;
        scanf("%lld%lld%lld%lld%lld", &n, &a, &b, &p, &q);
        ans += n / a * p;
        ans += n / b * q;
        ans -= n / (a * b / gcd(a, b)) * p;
        ans -= n / (a * b / gcd(a, b)) * q;
        if (p >= q)
            ans += n / (a * b / gcd(a, b)) * p;
        else
            ans += n / (a * b / gcd(a, b)) * q;
        printf("%lld\n", ans);
    }
    return 0;
}

G.单身狗的愤怒

#include <stdio.h>
#define N 1001
int n;
char s[N];
int cnt[26];
int main()
{
    scanf("%d",&n);
    scanf("%s",s);
    int mx = 0;
    for (int i=0;i<n;++i)
    {
        cnt[s[i]-'a'] ++;
        if (cnt[s[i]-'a']>mx) mx = cnt[s[i]-'a'];
    }
    if (mx>(n+1)/2) puts("NO");
    else puts("YES");
    return 0;
}

H.简单的计算

#include <stdio.h>
int main()
{
    long long f = 0, b = 0, tmp;
    for (int i = 0; i < 5; i++)
    {
        scanf("%lld", &tmp);
        f += tmp;
    }
    for (int i = 0; i < 5; i++)
    {
        scanf("%lld", &tmp);
        b += tmp;
    }
    printf("%lld", f * b);
    return 0;
}

I.Brain Power

#include <stdio.h>
#include <stdlib.h>
double score[][5] = {
    {1, 1, 0.8, 0.5, 0},
    {2, 2, 1.6, 1, 0},
    {3, 3, 2.4, 1.5, 0},
    {5, 5, 2.5, 2, 0},
    {1, 0.5, 0.4, 0.3, 0},
};
int num[5];
int main()
{
    double a0 = 0, a = 0, b0 = 0, b = 0, ans;
    for (int i = 0; i < 4; i++)
    {

        for (int j = 0; j < 5; j++)
            scanf("%d", &num[j]);
        for (int j = 0; j < 5; j++)
        {
            a += score[i][0] * num[j];
            a0 += score[i][j] * num[j];
        }

        if (i == 3)
        {
            for (int j = 0; j < 5; j++)
            {
                b += score[4][0] * num[j];
                b0 += score[4][j] * num[j];
            }
        }
    }
    ans = a0 / a * 100 + b0 / b;
    printf("%.8lf", ans);
    return 0;
}

J.掘地求升

#include <stdio.h>
#include <stdlib.h>
int main()
{
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    if (y < 0)
    {
        y = -y;
        x = -x;
        z = -z;
    }
    if (x < y)
        printf("%d", abs(x));
    else
    {
        if (z > y)
            printf("-1");
        else
            printf("%d", abs(z) + abs(x - z));
    }
    return 0;
}

K.我们是亚军(hard)

#include <stdio.h>

int main()
{

    int n;
    scanf("%d", &n);

    int mx1 = 0, id1, mx2 = 0, id2;

    for(int i = 1; i <= (1 << n); i ++)
    {
        int x;
        scanf("%d", &x);
        if(i <= (1 << (n - 1)))
        {
            if(x > mx1)
            {
                mx1 = x;
                id1 = i;
            }
        }
        else
        {
            if(x > mx2)
            {
                mx2 = x;
                id2 = i;
            }
        }
    }

    if(mx1 < mx2) printf("%d\n", id1);
    else printf("%d\n", id2);

    return 0;
}

L.我们是亚军(easy)

#include <stdio.h>

int a[(1 << 7) + 1], b[(1 << 7) + 1];

int id[(1 << 7) + 1];
int main()
{
    int n;
    scanf("%d", &n);
    int m = (1 << n);

    for(int i = 1; i <= m; i ++)
        scanf("%d", &a[i]), id[i] = a[i];
    
    for(int i = 1; i <= n - 1; i ++)
    {
        int cnt = 0;
        for(int j = 1; j <= 1 << (n - i + 1); j += 2)
            if(a[j] > a[j + 1])
                b[++ cnt] = a[j];
            else 
                b[++ cnt] = a[j + 1];
        
        for(int j = 1; j <= cnt; j ++)
            a[j] = b[j];
    }
    int t;
    if(a[1] < a[2]) t = a[1];
    else t = a[2];

    for(int i = 1; i <= m; i ++)
        if(t == id[i])
        {
            printf("%d\n", i);
            return 0;
        }
}

M.玩游戏

#include <stdio.h>

char s[4][110];
// int s[4][100];
int main()
{
    int t;
    scanf("%d", &t);

    while(t --)
    {
        int n;
        scanf("%d", &n);
        for(int i = 1; i <= 2; i ++)
            scanf("%s", s[i] + 1);

        int k = 0;
        for(int i = 1; i <= n; i ++)
        {
            if(s[1][i] == s[2][i] && s[1][i] == '1')
                k = 1;
        }
        if(k == 1) puts("NO");
        else puts("YES");
    }

    return 0;
}