project euler 16~20_projecteuler 20 未完成 n! 表示 n × (n

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本文深入探讨了Project Euler中四个经典问题的算法解决方案，包括大数运算、动态规划、日期计算及阶乘数字求和，提供了高效代码实现，是算法学习与竞赛准备的宝贵资源。

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Problem 16 Power digit sum

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思路

高精度乘法，这里用了万进制数，可以少一些循环

#include <iostream>
#include <cstdio>

using namespace std;

const int N = 1000 + 10;

int work(int x, int n) {
    int dig[N];
    int k = 1;
    dig[0] = 1;
    for(int i = 0; i < n; i++) {
        int carry = 0;
        for(int j = 0; j < k; j++) {
            carry += dig[j] * x;
            dig[j] = carry % 10000;
            carry /= 10000;
        }
        if(carry) {
            dig[k++] = carry;
        }
    }

    int ans = 0;
    for(int i = 0; i < k; i++) {
        while(dig[i] != 0) {
            ans += dig[i] % 10;
            dig[i] /= 10;
        }
    }
    return ans;
}

int main() {
    printf("%d\n", work(2, 1000));

    return 0;
}

Problem 17 Number letter counts

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#include <iostream>
#include <cstdio>
#include <map>

using namespace std;

map<int, string> mp;

void init() {
    mp[1] = "one";
    mp[2] = "two";
    mp[3] = "three";
    mp[4] = "four";
    mp[5] = "five";
    mp[6] = "six";
    mp[7] = "seven";
    mp[8] = "eight";
    mp[9] = "nine";
    mp[10] = "ten";
    mp[11] = "eleven";
    mp[12] = "twelve";
    mp[13] = "thirteen";
    mp[14] = "fourteen";
    mp[15] = "fifteen";
    mp[16] = "sixteen";
    mp[17] = "seventeen";
    mp[18] = "eighteen";
    mp[19] = "nineteen";
    mp[20] = "twenty";
    mp[30] = "thirty";
    mp[40] = "forty";
    mp[50] = "fifty";
    mp[60] = "sixty";
    mp[70] = "seventy";
    mp[80] = "eighty";
    mp[90] = "ninety";
    mp[100] = "hundred";
    mp[1000] = "thousand";
}

int work(int n) {
    int ans = 0;
    for(int i = 1; i <= n; i++) {
        int val = i;
        if(val >= 1000) {
            ans += mp[val/1000].size() + mp[1000].size();
            if(val % 1000 != 0) {
                ans += 3;
            }
            val %= 1000;
        }

        if(val >= 100) {
            ans += mp[val/100].size() + mp[100].size();
            if(val % 100 != 0) {
                ans += 3;
            }
            val %= 100;
        }

        if(val >= 20) {
            ans += mp[val-val%10].size();
            val %= 10;
        }

        if(val != 0) {
            ans += mp[val].size();
        }
    }
    return ans;
}

int main() {
    init();
    printf("%d\n", work(1000));

    return 0;
}

Problem 18 Maximum path sum I

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思路

动态规划，从上向下转移的话，状态转移方程 $d p [i] [j] + = m a x (d p [i - 1] [j - 1], d p [i - 1] [j])$ ，最终取最后一层中最大的值即可。当然也可以从下向上转移

#include <iostream>
#include <cstdio>

using namespace std;

const int N = 50;

int work(int dp[][N], int n) {
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= i; j++) {
            dp[i][j] += max(dp[i-1][j-1], dp[i-1][j]);
        }
    }

    int ans = 0;
    for(int i = 1; i <= n; i++) {
        ans = max(ans, dp[n][i]);
    }
    return ans;
}

int main() {
    int n = 15;
    int dp[N][N];
    memset(dp, 0, sizeof(dp));
    for(int i = 1; i <= 15; i++) {
        for(int j = 1; j <= i; j++) {
            scanf("%d", &dp[i][j]);
        }
    }

    printf("%d\n", work(dp, n));

    return 0;
}

Problem 19 Counting Sundays

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#include <iostream>
#include <cstdio>

using namespace std;

bool isLeap(int n) {
    return n%400 == 0 || (n%4==0 && n%100 != 0);
}

int work() {
    int month[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    int ans = 0, day = 1;
    for(int i = 1900; i <= 2000; i++) {
        for(int j = 1; j <= 12; j++) {
            if(i >= 1901 && day == 0) {
                ++ans;
            }

            if(j == 2) {
                day = (day + (isLeap(i) ? 29 : 28)) % 7;
            } else {
                day = (day + month[j]) % 7;
            }
        }
    }
    return ans;
}

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

    return 0;
}

Problem 20 Factorial digit sum

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思路

高精度乘法，这里用了万进制数

#include <iostream>
#include <cstdio>

using namespace std;

const int N = 1000;

int work(int n) {
    int dig[N];
    int k = 1;
    dig[0] = 1;

    for(int i = 1; i <= n; i++) {
        int carry = 0;
        for(int j = 0; j < k; j++) {
            carry += dig[j] * i;
            dig[j] = carry % 10000;
            carry /= 10000;
        }
        if(carry) {
            dig[k++] = carry;
        }
    }

    int ans = 0;
    for(int i = 0; i < k; i++) {
        while(dig[i] != 0) {
            ans += dig[i] % 10;
            dig[i] /= 10;
        }
    }
    return ans;
}

int main() {
    int n = 100;

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

    return 0;
}