project euler 21~25_project euler 25-优快云博客

本文链接：https://blog.youkuaiyun.com/jsszwc/article/details/103448689

本文集解析了多个经典算法问题，包括亲和数对的求和、名字得分问题、非盈数之和、字典序排列第N个排列及1000位斐波那契数的索引。通过高效算法实现，涵盖了数学、字符串处理和高精度计算等方面。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Problem 21 Amicable numbers

在这里插入图片描述

#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

const int N = 10000 + 10;

int work(int n) {
    int factorSum[N];
    for(int i = 1; i <= n; i++) {
        for(int j = i*2; j <= n; j += i) {
            factorSum[j] += i;
        }
    }

    int ans = 0;
    for(int i = 1; i < n; i++) {
        if(i >= factorSum[i]) {//避免重复取亲和数对
            continue;
        }
        if(factorSum[i] <= n && factorSum[factorSum[i]] == i) {
            ans += i + factorSum[i];
        }
    }
    return ans;
}

int main() {
    printf("%d\n", work(10000));

    return 0;
}

Problem 22 Names scores

在这里插入图片描述

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cctype>
#include <fstream>
#include <string>
#include <vector>

using namespace std;

typedef long long LL;

int score(string name) {
    int ans = 0;
    for(int i = 0; i < name.size(); i++) {
        ans += tolower(name[i]) - 'a' + 1;
    }
    return ans;
}

LL work(vector<string> &names) {
    sort(names.begin(), names.end());
    LL ans = 0;
    for(int i = 0; i < names.size(); i++) {
        ans += score(names[i]) * (i + 1);
    }
    return ans;
}

int main() {
    vector<string> names;
    ifstream fin("./../p022_names.txt");

    string str;
    fin >> str;
    int l = 0, r = 0;
    while(l < str.size()) {
        while(l < str.size() && str[l] != '\"') {
            l++;
        }
        r = l + 1;
        while(r < str.size() && isalpha(str[r])) {
            r++;
        }
        names.emplace_back(str.substr(l+1, r - l - 1));
        l = r + 1;
    }

    printf("%lld\n", work(names));

    return 0;
}

Problem 23 Non-abundant sums

在这里插入图片描述

思路：

用素数筛法的思想求出每个数字的真因子之和，于是根据每个数字的真因子之和就可以判断这个数字是不是盈数，然后用盈数两两求和标记可以表示成两个盈数的数字，剩下未标记的就是符合题目要求的数字了，计算其个数。

#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

const int N = 30000 + 10;

int work(int n) {
    int factSum[N];
    for(int i = 1; i <= n; i++) {
        for(int j = 2*i; j <= n; j += i) {
            factSum[j] += i;
        }
    }

    int arr[N];
    int k = 0;
    for(int i = 1; i <= n; i++) {
        if(factSum[i] > i) {
            arr[k++] = i;
        }
    }

    bool vis[N];
    fill(vis, vis + n + 1, false);
    for(int i = 0; i < k; i++) {
        for(int j = 0; j < k; j++) {
            if(arr[i] + arr[j] > n) {
                continue;
            }
            vis[arr[i]+arr[j]] = true;
        }
    }

    int ans = 0;
    for(int i = 0; i <= n; i++) {
        if(! vis[i]) {
            ans += i;
        }
    }

    return ans;
}

int main() {
    printf("%d\n", work(28123));

    return 0;
}

Problem 24 Lexicographic permutations

在这里插入图片描述

思路：

康托逆展开

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

using namespace std;

typedef long long LL;
const int N = 100 + 10;
LL fact[N];

void init(int n) {
    fact[0] = 1;
    for(int i = 1; i <= n; i++) {
        fact[i] = fact[i-1] * i;
    }
}

vector<int> work(vector<int> arr, LL k) {
    k -= 1;
    vector<int> ans;
    for(int i = arr.size(); i >= 1; i--) {
        int a = k / fact[i-1];
        k %= fact[i-1];
        ans.emplace_back(arr[a]);
        arr.erase(arr.begin() + a);
        sort(arr.begin(), arr.end());
    }
    return ans;
}

int main() {
    init(10);
    vector<int> arr = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
    vector<int> ans = work(arr, 1000000);
    for(int i = 0; i < ans.size(); i++) {
        printf("%d", ans[i]);
    }
    printf("\n");

    return 0;
}

Problem 25 1000-digit Fibonacci number

在这里插入图片描述

思路：

高精度加法求斐波那契数列

#include <iostream>
#include <cstdio>

using namespace std;

const int N = 1000 + 10;

void add(char *num1, char *num2, char *sum) {
    int len1 = strlen(num1), len2 = strlen(num2);
    int len = 0;
    reverse(num1, num1 + len1);
    reverse(num2, num2 + len2);

    int carry = 0;
    for(int i = 0; i < len1 || i < len2; i++) {
        carry += i < len1 ? num1[i]-'0' : 0;
        carry += i < len2 ? num2[i]-'0' : 0;
        sum[len++] = carry % 10 + '0';
        carry /= 10;
    }
    if(carry) {
        sum[len++] = carry + '0';
    }
    sum[len] = '\0';

    reverse(num1, num1 + len1);
    reverse(num2, num2 + len2);
    reverse(sum, sum + len);
}

int main() {
    char num1[N], num2[N], sum[N];
    num1[0] = '0', num1[1] = '\0';
    num2[0] = '1', num2[1] = '\0';

    int cnt = 1;
    while(true) {
        add(num1, num2, sum);
        ++cnt;
        if(strlen(sum) >= 1000) {
            break;
        }
        memcpy(num1, num2, sizeof(num2));
        memcpy(num2, sum, sizeof(sum));
    }

    printf("%d\n", cnt);

    return 0;
}