2018 年机考试题

1、求众数

#include <iostream>
#include <unordered_map>

using namespace std;

int main(){
    int n, num;
    cin >> n;
    int cnt = 0, ans = 0;
    unordered_map<int, int> m;
    for (int i = 0; i < n; i++) {
        cin >> num;
        m[num] ++;
        if (m[num] > cnt || m[num] == cnt && num < ans) ans = num;
    }
    cout << ans << endl;
    return 0;
}

2、解方程

#include <iostream>

using namespace std;

int main(){
    string str;
    cin >> str;
    int a = 0, b = 0, sym = 1, i = 0, flag = 1;//sym表示符号，flag=1表示等式左边，flag=-1表示等式右边
    while (i < str.size()) {
        if (str[i] == '=') {//遇到等号时，sym置1规避之前的影响
            flag = -1;
            sym = 1;
        }
        else if (str[i] == '+') sym = 1;
        else if (str[i] == '-') sym = -1;
        else {
            int t = 0;
            while (i < str.size() && str[i] >= '0' && str[i] <= '9') {
                t = 10 * t + (str[i] - '0');
                i++;
            }
            if (i >= str.size()) {//当表达式最后是数字时单独判断，如4x-1=2x+3，否则会越界
                b -= t * sym;
                break;
            }
            if (str[i] == 'x' && t == 0) a += sym * flag;//x前系数为正负1时单独判断处理
            else if (str[i] == 'x') a += t * sym * flag;//处理系数
            else {//处理常数
                b += t * sym * flag;
                continue;
            }
        }
        i++;
    }
    if (a == 0 && b == 0) printf("infinite solutions\n");
    else if (a != 0 && b % a == 0) printf("x=%d\n", -b / a);
    else printf("no solution\n");
    return 0;
}

3、骨牌

dp[i] = dp[i - 1] + dp[i - 2]

#include <iostream>

using namespace std;

const int MAXN = 10010;
int dp[MAXN];

int main() {
    int n;
    cin >> n;
    dp[1] = 1; dp[2] = 2;
    for (int i = 3; i <= n; i++) {
        dp[i] = dp[i - 1] + dp[i - 2];
    }
    cout << dp[n];
    return 0;
}