A.判断三角形

C++ 代码

#include <iostream>

using namespace std;

int main()
{
    int a, b, c;
    cin >> a >> b >> c;
    int sum = a + b + c;
    int max = a > b ? (a > c ? a : c) : (b > c ? b : c);
    int min = sum - max;
    if (max < min) puts("yes");
    else puts("no");
    return 0;    
}

B.超级楼梯

斐波那契数列

C++ 代码

#include <iostream>
#include <cstdio>

using namespace std;

const int N = 50;

int f[N];
int n, m;

int main()
{
    f[1] = 1;
    f[2] = 1;
    for (int i = 3; i < N; i ++ ) f[i] = f[i - 1] + f[i - 2];

    scanf("%d", &n);
    while (n -- )
    {
        scanf("%d", &m);
        cout << f[m] << endl;
    }
    return 0;
}

C.回文数猜想

C++ 代码

#include <iostream>
#include <cstdio>
#include <vector>

using namespace std;

vector<int> v;
int n, temp;

// 函数功能：判断参数n是否为回文数
// 参数n：需要判断是否为回文数的数
// 参数t：返回值，n的逆序数
bool isPalindrome(int n, int &t)
{
    int temp1 = n;

    t = 0;
    while (n > 0)
    {
        t = t * 10 + n % 10;
        n /= 10;
    }

    return temp1 == t;
}

int main()
{
    while (cin >> n)
    {
        v.clear();
        v.push_back(n);
        while (1)
        {
            if (isPalindrome(n, temp)) break;
            else
            {
                n += temp;
                v.push_back(n);
            }
        }

        cout << v.size() - 1 << endl;
        for (int i = 0; i < v.size(); i ++ )
        {
            if (i == 0) printf("%d", v[i]);
            else cout << "--->" << v[i];
        }
        puts("");
    }
    return 0;
}

D.最简单的计算机

C++ 代码

#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>

using namespace std;

int m1, m2, r1, r2, r3;

int main()
{
    while (cin >> m1 >> m2)
    {
        r1 = r2 = r3 = 0;
        string str;
        cin >> str;
        for (int i = 0; i < str.size(); i ++ )
        {
            char op = str[i];
            if (op == 'A') r1 = m1;
            else if (op == 'B') r2 = m2;
            else if (op == 'C') m1 = r3;
            else if (op == 'D') m2 = r3;
            else if (op == 'E') r3 = r1 + r2;
            else if (op == 'F') r3 = r1 - r2;
        }
        printf("%d,%d\n", m1, m2);
    }
    return 0;
}

E.钱币兑换问题

完全背包

C++ 代码

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 32768;

int v[] = {1, 2, 3};

int f[N];
int n;

int main()
{
    while (cin >> n)
    {
        memset(f, 0, sizeof f);

        f[0] = 1;
        for (int i = 0; i < 3; i ++ )
            for (int j = v[i]; j <= n; j ++ )
                f[j] += f[j - v[i]];

        cout << f[n] << endl;
    }
    return 0;
}

F.确定比赛名次

拓扑排序

C++ 代码

#include <iostream>
#include <cstring>
#include <queue>
#include <cstdio>
#include <functional>
#include <vector>
#include <algorithm>

using namespace std;

const int N = 510;

int n, m, cnt;
int h[N], e[N], ne[N], idx;
int d[N], ans[N]; // q[]存拓扑序列，d[]存每个点的入度

void add(int a, int b)  // 添加一条边a->b
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ;
}

void topsort()
{
    priority_queue<int, vector<int>, greater<int> > q;
    // 第一步寻找入度为零的点并插入队列
    for (int i = 1; i <= n; i ++ )
        if (!d[i])
            q.push(i);

    while (q.size())
    {
        int t = q.top(); // 每次取出队头判断
        q.pop();

        ans[cnt ++ ] = t;

        for (int i = h[t]; i != -1; i = ne[i])
        {
            int j = e[i];
            if (-- d[j] == 0)
                q.push(j);
        }
    }
}

int main()
{
    while (~scanf("%d%d", &n, &m))
    {
        memset(h, -1, sizeof h);
        memset(d, 0, sizeof d);
        idx = 0;
        cnt = 0;

        while (m -- )
        {
            int a, b;
            scanf("%d%d", &a, &b);
            add(a, b);
            d[b] ++ ; // b点的入度+1
        }
        topsort();
        for (int i = 0; i < cnt; i ++ )
        {
            if (i != cnt - 1) printf("%d ", ans[i]);
            else printf("%d\n", ans[i]);
        }
    }
    return 0;
}

G.找新朋友

欧拉函数

C++ 代码

#include <iostream>

using namespace std;

int phi(int x)
{
    int res = x;

    for (int i = 2; i <= x / i; i ++ )
        if (x % i == 0)
        {
            res = res / i * (i - 1);
            while (x % i == 0) x /= i;
        }

    if (x > 1) res = res / x * (x - 1);

    return res;
}

int main()
{
    int n;
    cin >> n;
    while (n -- )
    {
        int x;
        cin >> x;
        cout << phi(x) << endl;
    }
    return 0;
}

H.破译密码

isupper(int x) 检查x是否为大写字母(A ~ Z)。是，则返回1；不是，则返回0；

对于源码a，加密字母x，密文字母v，根据题意满足：a^x=v，那么有a^x^x=v^x，得a=v^x。对于所有的ｖ，用“A”到“Z”作为x去试算，满足v^x为“A”到“Z”即为加密字母x。

试探法寻找密码用大写字母是需要找到一个字母x，满足对于所有数据计算结果都为大写字母。

C++ 代码

#include <cstdio>
#include <iostream>
#include <cctype>

using namespace std;

const int N = 1010;

int n, v[N], x, j;

int main()
{
    while (cin >> n)
    {
        for (int i = 0; i < n; i ++ ) cin >> v[i];

        for (int i = 'A'; i <= 'Z'; i ++ ) // 试探法求加密字母x 
        {
            for (j = 0; j < n; j ++ )
                if (!isupper(v[j] ^ i)) // 密文字母v ^ 加密字母x = 原文字母a 
                    break;

            if (j == n) // 所有字母都为大写字母则满足题意 
            {
                x = i;
                break;
            }
        }

        for (int i = 0; i < n; i ++ ) printf("%c", v[i] ^ x);
        puts("");
    }
    return 0;
}

I.寒冰王座

完全背包

C++ 代码

#include <iostream>
#include <cstring>

using namespace std;

const int N = 10010;

int f[N];
int v[] = {150, 200, 350}, w[] = {150, 200, 350};

int max(int a, int b)
{
    return a > b ? a : b;
}

int main()
{
    int t;
    cin >> t;
    while (t -- )
    {
        memset(f, 0, sizeof f);
        int n;
        cin >> n;

        for (int i = 0; i < 3; i ++ )
            for (int j = v[i]; j <= n; j ++ )
                f[j] = max(f[j], f[j - v[i]] + w[i]);

        cout << n - f[n] << endl;
    }
    return 0;
}

J.Hat’s IEEE

一道float与double有区别的题目......

输入一个 32-bit 的数 n，输出指数和小数，满足 2^指数 × 小数 = 原数，并且小数的绝对值在 [1.0, 2.0) 的范围内。（注意 n 必须用 float 类型存储）

C++ 代码

#include <iostream>
#include <iomanip>
#include <cmath>

using namespace std;

int main()
{
    float n;
    int exponent; // 指数
    while (cin >> n)
    {
        exponent = 0;
        while (fabs(n) >= 2.0) 
        {
            n /= 2.0;
            exponent ++ ;
        }
        while (fabs(n) < 1.0)
        {
            n *= 2.0;
            exponent -- ; 
        }
        cout << exponent << ' ' << fixed << setprecision(6) << n << endl;
    }
    return 0; 
}