☆【容斥原理】【SCOI2010】幸运数字

【题目描述】在中国，很多人都把6和8视为是幸运数字！lxhgww也这样认为，于是他定义自己的“幸运号码”是十进制表示中只包含数字6和8的那些号码，比如68，666，888都是“幸运号码”！但是这种“幸运号码”总是太少了，比如在[1,100]的区间内就只有6个（6，8，66，68，86，88），于是他又定义了一种“近似幸运号码”。lxhgww规定，凡是“幸运号码”的倍数都是“近似幸运号码”，当然，任何的“幸运号码”也都是“近似幸运号码”，比如12，16，666都是“近似幸运号码”。现在lxhgww想知道在一段闭区间[a, b]内，“近似幸运号码”的个数。【输入】输入数据是一行，包括2个数字a和b【输出】输出数据是一行，包括1个数字，表示在闭区间[a, b]内“近似幸运号码”的个数【样例输入1】1 10【样例输出1】2【样例输入2】1234 4321【样例输出2】809【数据范围】   对于30%的数据，保证1<=a<=b<=1000000   对于100%的数据，保证1<=a<=b<=10000000000

这是一道使用容斥原理并剪枝优化的题。

首先找出所有的幸运数字的基数（即只含数字6和8的数），然后根据容斥原理的基本方法零壹枚举，具体细节如程序注释所示。
Accode：

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <bitset>using std::cin;using std::cout;typedef long long int64;const char fi[] = "luckynumber.in";const char fo[] = "luckynumber.out";const int maxN = 2050;int64 tmp[maxN];int64 Enum[maxN];int n, tot;int64 a, b, m;void init_file(){freopen(fi, "r", stdin);freopen(fo, "w", stdout);std::ios::sync_with_stdio(false);return;}void readdata(){    cin >> a >> b; --a;return;}void mknum(int64 p){    if (p > b) return;    if (p > 0) tmp[++tot] = p;    mknum(p * 10LL + 6LL);    mknum(p * 10LL + 8LL);    return;}int cmp(const void *a, const void *b){    if (*(int64 *)b > *(int64 *)a) return 1;    if (*(int64 *)b < *(int64 *)a) return -1;    return 0;} //对int64型的数排序只能这样。void modify(){    qsort(tmp + 1, tot, sizeof(tmp[0]), cmp);    for (int i = 1; i < tot + 1; ++i)    {        bool flag = true;        for (int j = tot; j > i; --j)        if (tmp[i] % tmp[j] == 0)        {            flag = false;            break;        }        if (flag) Enum[++n] = tmp[i];    }//将以上生成的基数中能被其它较小的基数整除的筛掉，//并从大到小排序，尽量较早地剪枝。    return;}inline int64 gcd(int64 a, int64 b){    int64 tmp;    while (b)    {        tmp = b;        b = a % b;        a = tmp;    }    return a;}int64 Dfs(int i, int64 num){    if (i > n) return b / num - a / num;    int64 ans = Dfs(i + 1, num);//若当前这个待枚举的数未被选，//则不改变枚举总个数的奇偶性，//就不翻转Dfs()前的符号。    int64 g = gcd(num, Enum[i]);    if (num / g <= b / Enum[i])        ans -= Dfs(i + 1, num / g * Enum[i]);//若Lcm(num, Enum[i]) <= b，则当前待枚举的数//可以被选，枚举时由于总个数增加了1，改变了//奇偶性，所以在Dfs()之前添加负号，就会形成//一层一层嵌套的负号的式子。    return ans;}void work(){    mknum(6LL); mknum(8LL);    modify();    cout << b - a - Dfs(1, 1) << std::endl;//输出最后结果的时候，记得Dfs()模块求出来的//是补集，需要对其取反。return;}int main(){init_file();readdata();work();return 0;}

第二次做：

#include <cstdio>typedef long long int64;const int64 Enum[] = 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a, b;inline int64 gcd(int64 a, int64 b){    while (b)    {int64 tmp = a % b; a = b; b = tmp;}    return a;}int64 Dfs(int i, int64 Last){    if (!Enum[i]) return b / Last - a / Last;    int64 tmp = Dfs(i + 1, Last);    int64 g = gcd(Last, Enum[i]);    if (Last / g <= b / Enum[i])        tmp -= Dfs(i + 1, Last / g * Enum[i]);    return tmp;}int main(){    freopen("luckynumber.in", "r", stdin);    freopen("luckynumber.out", "w", stdout);    scanf("%I64d%I64d", &a, &b); --a;    printf("%I64d\n", b - a - Dfs(0, 1));    return 0;}