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#include <bits/stdc++.h>
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using namespace std;
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/*
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题目：求0—7所能组成的奇数个数。 (可重复 0不能做首位)
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思路: 0-7能重复 统计1位,2位,3位, 4位, 5位, 6位,7位,8位,每个位数的奇数个数
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1位 4 （1，3，5，7）
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2位 7*4 =4*7 最后一位是4种，那么首位就是除了0，就是8-1=7个，乘法原理：7*4=28个
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3位 7*4*8
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4位 7*4*8*8
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5位 7*4*8*8*8
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6位 7*4*8*8*8*8
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7位 7*4*8*8*8*8*8
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8位 7*4*8*8*8*8*8*8
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*/
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int main() {
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//加一位和两位的个数
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int cnt = 4 + 4 * 7;
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int x = 8;
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//从3到8
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for (int i = 3; i <= 8; i++) {
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cnt += 7 * 4 * x;
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x = x * 8;
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}
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printf("%d\n", cnt);
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return 0;
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}
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