10月15日贝壳找房研发类试卷题解

全AC。
1、傻傻的搏斗

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 5;
typedef long long LL;

int main()
{
    LL T, X, A, C, B, D, Y;
    cin >> T;
    while (T--) {
        cin >> X >> A >> C >> Y >> B >> D;
        LL XIAOZHI = (X / B - 1) * D;  // 不用判断X是否小于等于B
        if (X % B != 0) XIAOZHI += D;

        LL XIAOCHUN = (Y / A - 1) * C;
        if (Y % A != 0) XIAOCHUN += C;

        if (XIAOCHUN == XIAOZHI) puts("TIE");
        else if (XIAOCHUN < XIAOZHI) puts("XIAOZHI");
        else puts("XIAOCHUN");
    }
    return 0;
}

2、简单的表达式计算

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 5;
typedef long long LL;

LL compute(int pos, string str, int n)
{
    LL len = str.size(), res = 0;
    for (int i = pos; i < len; i++) {
        if (str[i] >= 'A') res = res * n + str[i] - 'A' + 10;
        else res = res * n + str[i] - '0';
    }
    return res;
}

LL str2dec(string str)
{
    int len = str.size();
    if (len == 1) return str[0] - '0';
    if (len == 2) {
        if (str[0] == '0') return str[1] - '0';
        return (str[0] - '0') * 10 + str[1] - '0';
    }
    if (str[0] == '0' && str[1] == 'x') {
        return compute(2, str, 16);
    }
    else if (str[0] == '0') {
        return compute(1, str, 8);
    }
    return compute(0, str, 10);
}

int main()
{
    LL ans = 0;
    string str;
    cin >> str;
    int len = str.size();
    string s = "";
    int pre = 0;
    for (int i = 0; i < len; i++) {
        if (str[i] == '+') {
            LL tmp = str2dec(s);
            //cout << tmp << endl;
            s = "";
            if (!pre) ans += tmp;
            else ans -= tmp;
            pre = 0;
        }
        else if (str[i] == '-') {
            LL tmp = str2dec(s);
            //cout << tmp << endl;
            s = "";
            if (!pre) ans += tmp;
            else ans -= tmp;
            pre = 1;
        }
        else s += str[i];
    }
    LL tmp = str2dec(s);
    //cout << tmp << endl;
    if (!pre) ans += tmp;
    else ans -= tmp;
    cout << ans << endl;
    return 0;
}

3、找寻序列

#include <bits/stdc++.h>
using namespace std;

const int mod = 1e9 + 7;
typedef long long LL;
LL dp[1005][1005];

int main()
{
    for (int i = 1; i < 1005; i++) dp[1][i] = 1;
    for (int i = 2; i < 1005; i++) {      // 长度N
        for (int j = 1; j < 1005; j++) {  // 尾值M
            int tmp = sqrt(j);
            for (int k = 1; k <= tmp; k++) {
                if (j % k == 0) {
                    dp[i][j] += dp[i - 1][k];
                    dp[i][j] %= mod;
                    dp[i][j] += dp[i - 1][j / k];
                    dp[i][j] %= mod;
                }
            }
            if (tmp * tmp == j) {
                dp[i][j] -= dp[i - 1][tmp];
                dp[i][j] %= mod;
            }
        }
    }

    int N, M;
    while (cin >> N >> M) {
        cout << dp[N][M] << endl;
    }
    return 0;
}