杨帆班第7周代码

A 公式题

long long f[10010];
const int mod = 1e9 + 7;

void solve() {
    f[1] = 1;
    f[2] = 2;
    for (int i = 3; i <= 1000; i++) {
        f[i] = 2 * f[i - 1] + 3 * f[i - 2] + i; 
        f[i] %= mod;
        // (a + b) % mod = (a % mod + b % mod) % mod
    }
    int n;
    while (scanf("%d", &n), n) {
        printf("%d\n", f[n]);
    }
}
 

B 意大利青年的烦恼

 
char s[100010];
char res[100010];
 
void solve() {
    scanf("%s", s); 
    int idx = 0;
    for (int i = 0; s[i] != '\0'; i++) {
        if (s[i] == '4')
            continue;
        res[idx++] = s[i];
    }
    res[idx] = '\0';
    if (idx == 0)
        printf("YAMERO!\n");
    else
        printf("%s\n", res);
 
}

C 最大公约数

int gcd(int a, int b) {
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

void solve() {
    int a, b;
    scanf("%d %d", &a, &b);
    printf("%d\n", gcd(a, b));

}

D   Fibonacci数列

int f[1000010];
const int mod = 10007;
 
int func(int n) {
    if (f[n] != 0)
        return f[n];
    if (n == 1 || n == 2)
        return 1;
    int res = func(n - 1) + func(n - 2);
    res %= mod;
    f[n] = res;
    return res;
}
 
void solve() {
    f[1] = 1;
    f[2] = 1;
    int n;
    scanf("%d", &n);
    // (a + b) % mod = (a % mod + b % mod) %mod
    // for (int i = 3; i <= n; i++) {
    //     f[i] = f[i - 1] + f[i - 2];
    //     f[i] %= mod;
    // }
    printf("%d\n", func(n));
}

E 走楼梯

int f[100];
void solve() {
    f[1] = 1;
    f[2] = 2;
    int n;
    scanf("%d", &n);
    for (int i = 3; i <= n; i++) {
        f[i] = f[i - 1] + f[i - 2];
    } 
    printf("%d\n", f[n]);
}

F 杨辉三角

int f[100][100];
void solve() {
    int n;
    scanf("%d", &n); 
    f[1][1] = 1;
    for (int i = 2; i <= n; i++) {
        for (int j = 1; j <= i; j++) {
            f[i][j] = f[i - 1][j] + f[i - 1][j - 1];
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= i; j++) {
            printf("%d ", f[i][j]);
        }
        printf("\n");
    }
}

G 约数之和

void solve() {
    int n;
    scanf("%d", &n); 
    int res = 0;
    for (int i = 1; i <= n; i++) {
        if (n % i == 0) {
            res += i;
        }
    }
    printf("%d\n", res);
}

H 数的阶乘

void solve() {
    int n;
    scanf("%d", &n); 
    int res = 1;
    for (int i = 1; i <= n; i++) {
        res *= i; 
    }
    printf("%d\n", res);
}

I 数位五五

void solve() {
    int a, b;
    scanf("%d%d", &a, &b);
    int res = 0;
    for (int i = a; i <= b; i++) {
        int x = i;
        int sum = 0;
        while (x) {
            sum += x % 10;
            x /= 10; 
        }
        if (sum % 5 == 0)
            res++; 
    }
    printf("%d\n", res);
}

J The Biggest Water Problem

void solve() {
    int n;
    scanf("%d", &n);
    do {
        int sum = 0;
        while (n) {
            sum += n % 10;
            n /= 10;
        }
        n = sum;
    } while(n >= 10);
    printf("%d\n", n);
}

K 水仙花数

void solve() {
    int n;
    scanf("%d", &n);
    int nums[20], idx = 0;
    int t = n;
    while (n) {
        nums[idx++] = n % 10;
        n /= 10;
    }
    int res = 0;
    for (int i = 0; i < idx; i++) {
        int sum = 1;
        for (int j = 0; j < idx; j++) {
            sum *= nums[i];
        }
        res += sum;
    }
    if (res == t) {
        printf("yes\n");
    }
    else {
        printf("no\n");
    }
}

L   素数回文

int isPrime(long long n) {
    if (n <= 1)
        return 0;
    for (long long i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            return 0;
        }
    }
    return 1;
}
 
void solve() {
    int n;
    scanf("%d", &n);
    long long res = n;
    n /= 10;
    while (n) {
        res = res * 10 + n % 10;
        n /= 10;
    }
 
    if (isPrime(res))
        printf("prime\n");
    else
        printf("noprime\n");
}

M 数字统计

void solve() {
    int l, r;
    scanf("%d %d", &l, &r);
    int res = 0;
    for (int i = l;  i <= r; i++) {
        int x = i;
        while(x) {
            int t = x % 10;
            x /= 10;
            if (t == 2) {
                res++;
            }
        }
    }
    printf("%d\n", res);
}

N 递归次数

int cnt = 0;
int fact(int n) {
    cnt++;
    if (n <= 0) 
        return 1;
    return fact(n - 1) * n;
}
 
void solve() {
    int n;
    scanf("%d", &n);
    fact(n);
    printf("%d\n", cnt);
    // printf("%d\n", n + 1);
 
}

O 兔子的序列

#include <math.h>

void solve() {
    int n;
    scanf("%d", &n);
    int maxv = 0;
    for (int i = 0; i < n; i++) {
        int x;
        scanf("%d", &x);
        int t = sqrt(x); 
        if (t * t == x) {
            continue;
        }
        if (x > maxv) {
            maxv = x;
        }
    }
    printf("%d\n", maxv);
}