练习

A

#include <iostream>
#include <iomanip>
using namespace std;
double a, b, c, d;
double f(double x) {
    return a * x * x * x + b * x * x + c * x + d;
}
/*二分逼近*/
double find(double a, double b) {
    if (b - a < 0.0001) return a;
    double mid = (a + b) / 2;
    if (f(a) * f(mid) > 0) return find(mid, b);
    else find(a, mid);
}
int main() {
    double x;
    cin >> a >> b >> c >> d;
    for (double i = -100; i <= 100; i++) {
        if (f(i) == 0) cout << fixed << setprecision(2) << i << ' ';
        else if (f(i) * f(i + 1) < 0) cout << fixed << setprecision(2) << find(i, i + 1) << ' ';
    }
    return 0;
}

C++(看看有没有更省时的解法)

#include <iostream>
#include <cmath>
using namespace std;
int main() {
    int t, n;
    cin >> n;
    t = n;
    int round = 1;
    int sum = 1;
    while (t - sum > 0) {
        round++;
        sum += round;
    }
    int remain = round - abs(t - sum);
    if (round % 2) {
        cout << round - remain + 1 << "/" << remain;
    }
    else {
        cout << remain<< "/" << round - remain + 1;
    }
    return 0;
}

E

#include <iostream>
using namespace std;
bool yes_indeed(int a, int b, int c) {
    int v[10] = { 0 };
    while (a) {
        v[a % 10] = 1;
        a /= 10;
    }
    while (b) {
        v[b % 10] = 1;
        b /= 10;
    }
    while (c) {
        v[c % 10] = 1;
        c /= 10;
    }
    for (int i = 1; i <= 9; i++) {
        if (v[i] == 0) return 0;
    }
    return 1;
}
int main() {
    int a, b, c;
    for (int a = 123; a <= 329; a++) {
        b = 2 * a;
        c = 3 * a;
        if (yes_indeed(a, b, c)) cout << a << ' ' << b << ' ' << c << endl;
    }
    return 0;
}

F(来自洛谷，我自己做的超时GG了)

#include<iostream>
#include<cmath>
using namespace std;
typedef long long ll;
int m, n, ans, flag;
ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a % b);
}
int main()
{
    cin >> n >> m;
    for (int i = 1; i <= sqrt(1ll * m * n); i++)
    {
        if ((1ll * n * m) % i == 0 && gcd(i, (1ll * n * m) / i) == n)
        {
            ans++;
            if (1ll * i * i == 1ll * n * m)  flag = 1;  //两个数相同的情况
        }
    }
    cout << ans * 2 - flag;
    return 0;
}

G - 连续自然数和(暴力出奇迹，看看怎么用前缀解题)

#include<iostream>
using namespace std;
int n, sum, j;
int main()
{
    cin >> n;         
    for (int i = 1; i <= n / 2; i++)
    {
        sum = 0;     
        for (j = i; j < n; j++)   
        {
            sum += j;            
            if (sum >= n)break;   
        }
        if (sum == n)cout << i << ' ' << j << endl;   
    }
    return 0;
}

H - 约瑟夫

https://blog.csdn.net/u012135009/article/details/99443183