Saruman’s Level Up~（多校赛算组合数）

最新推荐文章于 2021-06-02 14:57:38 发布

转载最新推荐文章于 2021-06-02 14:57:38 发布 · 88 阅读

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原文链接：http://www.cnblogs.com/qldabiaoge/p/8877867.html

本文介绍了一个基于二进制表示中1的数量为3的倍数的天数来预测萨鲁曼塔楼等级的问题。通过将给定的天数转换为二进制并计算符合条件的天数来确定塔楼的等级。

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Description

Saruman’s army of orcs and other dark minions continuously mine and harvest lumber out of the land surrounding his mighty tower for N continuous days. On day number i, Saruman either chooses to spend resources on mining coal and harvesting more lumber, or on raising the level (i.e., height) of his tower. He levels up his tower by one unit only on days where the binary representation of i contains a total number of 1’s that is an exact multiple of 3. Assume that the initial level of his tower on day 0 is zero. For example, Saruman will level up his tower on day 7 (binary 111), next on day 11 (binary 1011) and then day 13, day 14, day 19, and so on. Saruman would like to forecast the level of his tower after N days. Can you write a program to help?

Input

The input file will contain multiple input test cases, each on a single line. Each test case consists of a positive integer N < 1016, as described above. The input ends on end of file.

Output

For each test case, output one line: “Day N: Level = L”, where N is the input N, and L is the number of levels after N days.

Sample Input

2
19
64

Sample Output

Day 2: Level = 0
Day 19: Level = 5
Day 64: Level = 21


这题其实就是一个模拟，其实把n转化为二进制，
然后求出小于等于n有多少个数的二进制中含有1的个数是3的倍数
就是计算组合数，思维题，
例如1000100  当计算到第一个1的时候 就是加上c(6,3)+c（6，6）
此时bits的个数加一，然后到第二个1的时候就是看看
（i+bits)%==0 0<=i<=2 时候存在 i 再加上组合数就是了

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <queue>
 5 typedef long long LL;
 6 LL gcd(LL a, LL b ) {
 7     if (b == 0) return a;
 8     return gcd(b, a % b);
 9 }
10 LL combine(LL n, LL r) {
11     LL ret = 1;
12     if (r > n - r) r = n - r;
13     for (LL i = 1 ; i <= r ; i++) {
14         LL k = n - i + 1;
15         LL g = gcd(i, k);
16         ret = ret / (i / g) * (k / g);
17     }
18     return ret;
19 }
20 LL ans(LL n) {
21     LL ret = 0, bits = 0, k = ((LL)1 << 60), r = 60;
22     while(k) {
23         if (k & n) {
24             for (int i = 0 ; i <= r ; i++) {
25                 if ((i + bits) > 0 && ((i + bits) % 3 == 0)) ret += combine(r, i);
26             }
27             bits++;
28         }
29         k = k >> 1;
30         r--;
31     }
32     if (bits > 0 && bits % 3 == 0) ret++;
33     return ret;
34 }
35 int main() {
36     LL n;
37     while(scanf("%lld", &n) != EOF) {
38         printf("Day %lld: Level = %lld\n", n, ans(n));
39     }
40     return 0;
41 }

转载于:https://www.cnblogs.com/qldabiaoge/p/8877867.html