原始题目

P1940 Reversible Number

题目大意

有些正整数n可能满足n + 回文(n)（回文(n)是把n倒过来写所得的数）得到的结果的各位都是奇数。那么，小于等于10^x的Reversible数有多少个？方便起见，x是大于等于3小于等于400的正整数。

解题思路

很明显不可能是暴力的，题目范围给400位，所以暗示我们跟位数有关系。

暴力求前几位能得到下面数据

1-9 0

10-99 20

100-999 100

1000-9999 600

10000-99999 0

100000-999999 18000

1000000-9999999 50000

发现奇偶位是有规律的。

数学如下

1. 偶数位数；

   若要保证回文性质，要求全部位不能进位

   可用反证法证明，推矛盾,所以有:

   \[ even[2] = 20 , even[2n] = even[2(n-1)] * 30\]

2. 奇数位数：

   必然是 进位|不进位|进位

   进位|不进位|进位|不进位|进位|不进位|进位

   即模4余3位数时可以找到满足的reversible数。

   \[ odd[3] = [100], odd[4n+3] = odd[4(n-1)+3] * 500 \]

暴力对拍代码

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

#define rep(i, a, n) for (ll i = a; i < n; ++i)
#define per(i, a, n) for (ll i = n - 1; i >= a; --i)

ll n, m, N;
int main()
{
    while (cin >> n >> m >> N) {
        ll cnt = 0;
        rep(i, n, m + 1)
        {
            ll flag = 1;
            if (i % 10 == 0)
                continue;
            ll temp = i;
            ll sum = temp;
            for (ll j = 0; j < N; j++) {
                ll addsum = temp % 10;
                for (ll k = 0; k < N - j - 1; k++)
                sum += addsum;
                temp /= 10;
            }
            temp = sum;
            while (temp) {
                if ((temp & 1) == 0)
                    flag = 0;
                temp /= 10;
            }
            // cout << sum << endl;
            if (flag) {
                // cout << "i=" << i << "  sum=" << sum << endl;
                cnt++;
            }
        }
        cout << cnt << endl;
    }
}

解题代码

// 1-9 0
// 10-99 20
// 100-999 100
// 1000-9999 600
// 10000-99999 0
// 100000-999999 18000
// 1000000-9999999 50000 ???
//

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

#define rep(i, a, n) for (int i = a; i < n; ++i)
#define per(i, a, n) for (int i = n - 1; i >= a; --i)
#define fi first
#define se second
#define pb push_back
#define np next_permutation
#define INF 0x3f3f3f3f
#define EPS 1e-8
#define eps '\n'

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<string, string> pss;
typedef vector<int> vi;

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <string>
using namespace std;

#define MAXN 9999
#define MAXSIZE 10
#define DLEN 4

class BigNum {
private:
    int a[500]; //可以控制大数的位数
    int len; //大数长度
public:
    BigNum()
    {
        len = 1;
        memset(a, 0, sizeof(a));
    } //构造函数
    BigNum(const int); //将一个int类型的变量转化为大数
    BigNum(const char*); //将一个字符串类型的变量转化为大数
    BigNum(const BigNum&); //拷贝构造函数
    BigNum& operator=(const BigNum&); //重载赋值运算符，大数之间进行赋值运算

    friend istream& operator>>(istream&, BigNum&); //重载输入运算符
    friend ostream& operator<<(ostream&, BigNum&); //重载输出运算符

    BigNum operator+(const BigNum&) const; //重载加法运算符，两个大数之间的相加运算
    BigNum operator-(const BigNum&) const; //重载减法运算符，两个大数之间的相减运算
    BigNum operator*(const BigNum&)const; //重载乘法运算符，两个大数之间的相乘运算
    BigNum operator/(const int&) const; //重载除法运算符，大数对一个整数进行相除运算

    BigNum operator^(const int&) const; //大数的n次方运算
    int operator%(const int&) const; //大数对一个int类型的变量进行取模运算
    bool operator>(const BigNum& T) const; //大数和另一个大数的大小比较
    bool operator>(const int& t) const; //大数和一个int类型的变量的大小比较

    void print(); //输出大数
};
BigNum::BigNum(const int b) //将一个int类型的变量转化为大数
{
    int c, d = b;
    len = 0;
    memset(a, 0, sizeof(a));
    while (d > MAXN) {
        c = d - (d / (MAXN + 1)) * (MAXN + 1);
        d = d / (MAXN + 1);
        a[len++] = c;
    }
    a[len++] = d;
}
BigNum::BigNum(const char* s) //将一个字符串类型的变量转化为大数
{
    int t, k, index, l, i;
    memset(a, 0, sizeof(a));
    l = strlen(s);
    len = l / DLEN;
    if (l % DLEN)
        len++;
    index = 0;
    for (i = l - 1; i >= 0; i -= DLEN) {
        t = 0;
        k = i - DLEN + 1;
        if (k < 0)
            k = 0;
        for (int j = k; j <= i; j++)
            t = t * 10 + s[j] - '0';
        a[index++] = t;
    }
}
BigNum::BigNum(const BigNum& T)
    : len(T.len) //拷贝构造函数
{
    int i;
    memset(a, 0, sizeof(a));
    for (i = 0; i < len; i++)
        a[i] = T.a[i];
}
BigNum& BigNum::operator=(const BigNum& n) //重载赋值运算符，大数之间进行赋值运算
{
    int i;
    len = n.len;
    memset(a, 0, sizeof(a));
    for (i = 0; i < len; i++)
        a[i] = n.a[i];
    return *this;
}
istream& operator>>(istream& in, BigNum& b) //重载输入运算符
{
    char ch[MAXSIZE * 4];
    int i = -1;
    in >> ch;
    int l = strlen(ch);
    int count = 0, sum = 0;
    for (i = l - 1; i >= 0;) {
        sum = 0;
        int t = 1;
        for (int j = 0; j < 4 && i >= 0; j++, i--, t *= 10) {
            sum += (ch[i] - '0') * t;
        }
        b.a[count] = sum;
        count++;
    }
    b.len = count++;
    return in;
}
ostream& operator<<(ostream& out, BigNum& b) //重载输出运算符
{
    int i;
    cout << b.a[b.len - 1];
    for (i = b.len - 2; i >= 0; i--) {
        cout.width(DLEN);
        cout.fill('0');
        cout << b.a[i];
    }
    return out;
}

BigNum BigNum::operator+(const BigNum& T) const //两个大数之间的相加运算
{
    BigNum t(*this);
    int i, big; //位数
    big = T.len > len ? T.len : len;
    for (i = 0; i < big; i++) {
        t.a[i] += T.a[i];
        if (t.a[i] > MAXN) {
            t.a[i + 1]++;
            t.a[i] -= MAXN + 1;
        }
    }
    if (t.a[big] != 0)
        t.len = big + 1;
    else
        t.len = big;
    return t;
}
BigNum BigNum::operator-(const BigNum& T) const //两个大数之间的相减运算
{
    int i, j, big;
    bool flag;
    BigNum t1, t2;
    if (*this > T) {
        t1 = *this;
        t2 = T;
        flag = 0;
    } else {
        t1 = T;
        t2 = *this;
        flag = 1;
    }
    big = t1.len;
    for (i = 0; i < big; i++) {
        if (t1.a[i] < t2.a[i]) {
            j = i + 1;
            while (t1.a[j] == 0)
                j++;
            t1.a[j--]--;
            while (j > i)
                t1.a[j--] += MAXN;
            t1.a[i] += MAXN + 1 - t2.a[i];
        } else
            t1.a[i] -= t2.a[i];
    }
    t1.len = big;
    while (t1.a[t1.len - 1] == 0 && t1.len > 1) {
        t1.len--;
        big--;
    }
    if (flag)
        t1.a[big - 1] = 0 - t1.a[big - 1];
    return t1;
}

BigNum BigNum::operator*(const BigNum& T) const //两个大数之间的相乘运算
{
    BigNum ret;
    int i, j, up;
    int temp, temp1;
    for (i = 0; i < len; i++) {
        up = 0;
        for (j = 0; j < T.len; j++) {
            temp = a[i] * T.a[j] + ret.a[i + j] + up;
            if (temp > MAXN) {
                temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);
                up = temp / (MAXN + 1);
                ret.a[i + j] = temp1;
            } else {
                up = 0;
                ret.a[i + j] = temp;
            }
        }
        if (up != 0)
            ret.a[i + j] = up;
    }
    ret.len = i + j;
    while (ret.a[ret.len - 1] == 0 && ret.len > 1)
        ret.len--;
    return ret;
}
BigNum BigNum::operator/(const int& b) const //大数对一个整数进行相除运算
{
    BigNum ret;
    int i, down = 0;
    for (i = len - 1; i >= 0; i--) {
        ret.a[i] = (a[i] + down * (MAXN + 1)) / b;
        down = a[i] + down * (MAXN + 1) - ret.a[i] * b;
    }
    ret.len = len;
    while (ret.a[ret.len - 1] == 0 && ret.len > 1)
        ret.len--;
    return ret;
}
int BigNum::operator%(const int& b) const //大数对一个int类型的变量进行取模运算
{
    int i, d = 0;
    for (i = len - 1; i >= 0; i--) {
        d = ((d * (MAXN + 1)) % b + a[i]) % b;
    }
    return d;
}
BigNum BigNum::operator^(const int& n) const //大数的n次方运算
{
    BigNum t, ret(1);
    int i;
    if (n < 0)
        exit(-1);
    if (n == 0)
        return 1;
    if (n == 1)
        return *this;
    int m = n;
    while (m > 1) {
        t = *this;
        for (i = 1; i << 1 <= m; i <<= 1) {
            t = t * t;
        }
        m -= i;
        ret = ret * t;
        if (m == 1)
            ret = ret * (*this);
    }
    return ret;
}
bool BigNum::operator>(const BigNum& T) const //大数和另一个大数的大小比较
{
    int ln;
    if (len > T.len)
        return true;
    else if (len == T.len) {
        ln = len - 1;
        while (a[ln] == T.a[ln] && ln >= 0)
            ln--;
        if (ln >= 0 && a[ln] > T.a[ln])
            return true;
        else
            return false;
    } else
        return false;
}
bool BigNum::operator>(const int& t) const //大数和一个int类型的变量的大小比较
{
    BigNum b(t);
    return *this > b;
}

void BigNum::print() //输出大数
{
    int i;
    cout << a[len - 1];
    for (i = len - 2; i >= 0; i--) {
        cout.width(DLEN);
        cout.fill('0');
        cout << a[i];
    }
    cout << endl;
}

int n, m;
int main()
{

    while (~scanf("%d", &n)) {
        BigNum ans;
        if (n == 1)
            printf("0\n");
        else if (n == 2)
            printf("20\n");
        else if (n == 3) {
            printf("120\n");
        }
        // continue;
        else {
            ans = 120;
            BigNum tempeven = 20;
            BigNum tempodd = 100;
            for (int i = 4; i <= n; i++) {
                if (i % 4 == 3) {
                    ans = ans + (tempodd = tempodd * 500);
                }

                if ((i & 1) == 0) {
                    ans = ans + (tempeven = tempeven * 30);
                }
            }
            ans.print();
        }
    }
}

收获与反思

奇偶位发现规律并且找出可能数情况的本质，而不是单纯对数据找规律。