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UVa 1645 Count (递推,数论)
题意:给定一棵 n 个结点的有根树,使得每个深度中所有结点的子结点数相同。求多棵这样的树。
析:首先这棵树是有根的,那么肯定有一个根结点,然后剩下的再看能不能再分成深度相同的子树,也就是说是不是它的约数。那么答案就有了,
我们只要去计算n-1的约数有多少棵不同的树,然后就有递推式了。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const LL LNF = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1e3 + 5;const int mod = 1e9 + 7;const int dr[] = {-1, 0, 1, 0};const int dc[] = {0, 1, 0, -1};const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m;}LL dp[maxn];void init(){ dp[1] = 1; dp[2] = 1; for(int i = 2; i < 1000; ++i){ for(int j = 1; j <= i; ++j) if(i % j == 0) dp[i+1] = (dp[i+1] + dp[j]) % mod; }}int main(){ init(); int kase = 0; while(scanf("%d", &n) == 1){ printf("Case %d: %lld\n", ++kase, dp[n]); } return 0;}
UVa 1645 Count (递推,数论)
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