更偏向编程

值得一看

D 字符串匹配

#include <iostream>
#include <string>
using namespace std;
int main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int T;
    cin >> T;
    cin.get();
    while (T--) {
        string word, sentence;
        getline(cin, word);
        getline(cin, sentence);
        /*全部化为小写*/
        for (string::iterator i1 = word.begin(); i1 != word.end(); i1++)
            *i1 = tolower(*i1);
        for (string::iterator i2 = sentence.begin(); i2 != sentence.end(); i2++)
            *i2 = tolower(*i2);
        /*关键：去除局部匹配*/
        word = ' ' + word + ' ';
        sentence = ' ' + sentence + ' ';
        int i = sentence.find(word);
        int position = i;  //第一次出现的位置
        int cnt = 0;
        while (i != string::npos) {
            i = sentence.find(word, i + 1);  //从i+1开始匹配
            cnt++;
        }
        if (cnt == 0)
            cout << -1 << endl;
        else
            cout << cnt << " " << position << endl;
    }
    return 0;
}

H 全排列

一个仅由0-9组成的字符串，对于它所有排列所构成的数字按升序排列后的序列，第n个数是多少呢？

#include <bits/stdc++.h>
#include <sstream>
using namespace std;
typedef long long ll;
const int N = 1e5 + 7;
const ll mod = 1e9 + 7;
int main() {
    int T;
    cin >> T;
    for (int ca = 1; ca <= T; ++ca) {
        string a;
        int n;
        cin >> a >> n;
        int len = a.length();
        int cnt = 0;
        cout << "Case " << ca << ": ";
        sort(a.begin(),a.end());
        do {
            cnt++;
            if (cnt == n) {
                int ans = 0;
                stringstream s;
                s << a;
                s >>ans;
                cout << ans << endl;
                break;
            }
        } while (next_permutation(a.begin(), a.end()));
        if (cnt < n) cout << "no solution" << endl;
    }
    return 0;
}

string的一些技巧
https://blog.csdn.net/xiong452980729/article/details/61677701

G 数学题

本题可打表。

通过倍数递增的循环完成f数组的预处理，然后推过去就可以了。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
inline ll read() {
    ll s = 0, f = 1;
    char ch;
    do {
        ch = getchar();
        if (ch == '-') f = -1;
    } while (ch < 48 || ch > 57);
    while (ch >= 48 && ch <= 57)
        s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
    return s * f;
}
const int MOD = 1e9 + 7;
const int N = 1e6 + 5;
ll sum[N];
void init() {
    fill(sum, sum + N, 1);
    for (int i = 1; i < N; ++i)//此处处理为点睛之笔
        for (int j = i; j < N; j += i) sum[j] = sum[j] * i % MOD;
}
int main() {
    init();
    int T = read();
    while (T--) {
        int l = read(), r = read();
        ll ans = 1;
        for (int i = l; i <= r; ++i) ans = (ans * sum[i]) % MOD;
        printf("%lld\n", ans);
    }
    return 0;
}

水题

A 模拟

t=int(input())
for _ in range(t):
    n=int(input())
    ans=0
    while n!=1:
        if n%2 :
            n-=1
            ans+=1
        n//=2
        ans+=1
    print(ans)

B 字符串key-value

dict={'茕': 'qiong','孑': 'jie','立': 'li','沆': 'hang','瀣': 'xie','一': 'yi','气': 'qi','踽':'ju','独':'du','行': 'xing','醍': 'ti','醐': 'hu','灌': 'guan','顶':'ding','绵':'mian','瓜': 'gua','瓞': 'die','奉': 'feng','为': 'wei','圭': 'gui','臬': 'nie'}
for _ in range(int(input())): print(dict[input()])

C 坑题

给你一组数据，取出任意个数，使得他们的和相加最大且为奇数。

for _ in range(int(input())):
    n = int(input())
    a = list(map(int, input().split()))
    sum = 0
    for i in a:
        if i > 0:sum += i
    if sum & 1==0:
        a=sorted(a, key=abs)
        for i in a:
            if i&1:
                sum-=abs(i)
                break
    print(sum)

I GPArank

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
double eps = 1e-7;
struct Node {
    string id;
    double score;
};
vector<Node> p;
int a[15];
int b[4];
bool cmp(Node a, Node b) {
    if (fabs(a.score - b.score) > eps) return a.score > b.score;
    return a.id < b.id;
}
int main() {
    int T;
    cin >> T;
    while (T--) {
        p.clear();
        int n, m;
        cin >> n >> m;
        double sum = 0.0;
        for (int i = 1; i <= m; ++i) cin >> a[i], sum += a[i];
        for (int i = 1; i <= n; ++i) {
            string temp;
            cin >> temp;
            double sc = 0.0;
            for (int j = 1; j <= m; ++j) {
                for (int k = 1; k <= 3; ++k) cin >> b[k];
                sc += (b[1] * 0.5 + b[2] * 0.3 + b[3] * 0.2) * a[j];
            }
            p.push_back(Node{temp, sc / sum});
        }
        stable_sort(p.begin(), p.end(), cmp);
        for (int i = 0; i < n; ++i) {
            cout << p[i].id;
            printf(" %.2f\n", p[i].score);
        }
        if (T) printf("------\n");
    }
    return 0;
}