标签：ble printf list 技术分享 deque queue tor sig bsp
题意：已知
，可得出 P(1) = 4, P(2) = 1, P(3) = 5，由此可得出 P(P(1)) = P(4) = 2. And P(P(3)) = P(5) = 3，因此
。经过k次如上变换，最终可得
，输入保证一定有解，求k。
分析：
1、能用数组表示映射就别用map，很耗时
2、序列中的每个数字经过这种变换都一定会变会自身，例如，一开始P(1) = 4，接下来P(P(1)) = P(4) = 2,再接下来P(P(P(1))) = P(P(4)) = P(2) = 1,只需三次，就可以变回自身（P(P(P(1))) =1），对序列中每个数字求次数，最终求这些次数的最小公倍数即可
#include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<stack> #include<deque> #include<queue> #include<list> #define Min(a, b) a < b ? a : b #define Max(a, b) a < b ? b : a typedef long long ll; typedef unsigned long long llu; const int INT_INF = 0x3f3f3f3f; const int INT_M_INF = 0x7f7f7f7f; const ll LL_INF = 0x3f3f3f3f3f3f3f3f; const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f; const int dr[] = {0, 0, -1, 1}; const int dc[] = {-1, 1, 0, 0}; const double pi = acos(-1.0); const double eps = 1e-8; const int MAXN = 1000 + 10; const int MAXT = 10000 + 10; using namespace std; int x[MAXN]; int w[MAXN]; set<int> y; int gcd(int a, int b){ return !b ? a : gcd(b, a % b); } int N; int main() { scanf("%d", &N); for(int i = 1; i <= N; ++i) { scanf("%d", &x[i]); w[i] = x[i]; } for(int i = 1; i <= N; ++i) { int cnt = 1; while(x[i] != i){ x[i] = w[x[i]]; ++cnt; } y.insert(cnt); } set<int>::iterator it = y.begin(); int ans = *it; ++it; for(; it != y.end(); ++it){ ans = ans / gcd(ans, *it) * (*it);//求最小公倍数，调用函数耗时 } printf("%d\n", ans); return 0; }
标签：ble printf list 技术分享 deque queue tor sig bsp
原文地址：http://www.cnblogs.com/tyty-Somnuspoppy/p/6032976.html
