蓝桥杯基础题集训练一

蓝桥杯基础题集训练一

目录

• 1、字母图形
• 2、01字串
• 3、Fibonacci数列
• 4、圆的面积
• 5、十六进制转八进制
• 6、十六进制转十进制
• 7、十进制转十六进制

1、字母图形

//a ber
#include 
using namespace std;
int n, m;
char a[100][100];
signed main() {
    scanf("%d%d", &n, &m);
    //打印第一行
    for(int i = 0; i < m; i++) {
        a[0][i] = i + 'A';
    }

    for(int i = 1; i <= n; i++) {
        for(int j = 0; j < m; j++) {
            //规律
            i > j ? a[i][j] = char(a[i - 1][j] + 1) : a[i][j] = char(a[i - 1][j] - 1);
        }
    }
    for(int i = 0; i < n; i++) {
        puts(a[i]);
    }

    return 0;
}

2、01字串

//a ber
//水题 当成二进制数做就行了 0-31
#include 
using namespace std;
int a[5];
void print(int x) {
    memset(a, 0, sizeof a);
    int cnt = 0;
    while(x) {
        int c = x % 2;
        a[cnt++] = c;
        x /= 2;
    }
    a[cnt++] = x;
    for(int i = 4; i >= 0; i--) printf("%d", a[i]);
}
signed main() {
    for(int i = 0; i < 32; i++) {
        print(i);
        puts("");
    }
    return 0;
}

3、Fibonacci数列

//a ber
#include 
using namespace std;
const int N = 1e6 + 7;
int n, f[N];
signed main() {
    scanf("%d", &n);
    f[1] = f[2] = 1;
    if(n == 1 || n == 2) {
        puts("1");
        return 0;
    }

    for(int i = 3; i <= n; i++) {
        f[i] = (f[i - 1] + f[i - 2]) % 10007;
    }

    printf("%d",f[n]);

    return 0;
}

4、圆的面积

acos(-1) 圆周率

//a ber
#include 
using namespace std;
const double pi  = acos(-1);
double x;
signed main() {
    scanf("%lf",&x);
    printf("%.7f",pi*x*x);
    return 0;
}

5、十六进制转八进制

将十六进制转化成二进制，再将二进制转化成八进制，注意前导0处理

#include 
using namespace std;
string s;
string a[16] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001",
                "1010", "1011", "1100", "1101", "1110", "1111"
               };
map ok;

signed main() {
    ok["000"] = "0";
    ok["001"] = "1";
    ok["010"] = "2";
    ok["011"] = "3";
    ok["100"] = "4";
    ok["101"] = "5";
    ok["110"] = "6";
    ok["111"] = "7";
    int _;
    scanf("%d", &_);

    for(int T = 0; T < _; T++) {
        cin >> s;
        int n = s.size();
        string b;
        for(int i = 0; i < n; i++) {
            s[i] > '9' ? b += a[s[i] - 'A' + 10] : b += a[s[i] - '0'];
        }

        int nn =  b.size();
        //处理位数不足三的倍数的情况
        if(nn % 3 == 1) {
            b = "00" + b;
        }
        if(nn % 3 == 2) {
            b = "0" + b;
        }
        int f = 0;
        for(int j = 0; j < nn; j += 3) {
            string tmp = b.substr(j, 3);
            string t = ok[tmp];
            //前导0的处理
            if(j == 0 && t == "0") {
                f = 1;
                continue;
            }
            if(f == 1 && j == 3 && t == "0") {
                continue;
            }
            cout << t;
        }
        cout << "n";
    }

    return 0;
}

6、十六进制转十进制

#include 
#define int long long
using namespace std;
string s;
int fp(int a, int b) {
    int ok = 1;
    while(b) {
        if(b & 1) ok *= a;
        a *= a;
        b >>= 1;
    }
    return ok;
}
signed main() {
    int ok = 0;
    cin >> s;
    reverse(s.begin(), s.end());


    for(int _ = 0; _ < s.size(); _++) {
        s[_] > '9' ? ok = ok + (s[_] - 'A' + 10) * fp(16, _) : ok = ok + (s[_] - '0') * fp(16, _);
    }
    printf("%lld", ok);
    return 0;
}

7、十进制转十六进制

除k取余法emmm。。。

#include 
using namespace std;
int a;
vector ok;
signed main() {
    int a;
    scanf("%d", &a);
    int tmp;
    if(a == 0) {
        puts("0");
        return 0;
    }


    while(a) {
        tmp = a % 16;
        tmp > 9 ? ok.push_back(char('A' + tmp - 10)) : ok.push_back(char('0' + tmp));
        a /= 16;
    }

    for(int _ = ok.size() - 1; _ >= 0; _--) {
        printf("%c", ok[_]);
    }
    return 0;
}