题意:约瑟夫环,求最后三个数。
析:f[i] = (f[i-1] + k) % i 这是求最后一个数时候,我们倒着推到第一个数时,只有一个数,所以当只有两个数时,就是另一数,
同理,我们可以求得第三个数。
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> #include <unordered_map> #include <unordered_set> #define debug() puts("++++"); #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 double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 5e5 + 5; const int mod = 2000; const int dr[] = {-1, 1, 0, 0}; const int dc[] = {0, 0, 1, -1}; const char *de[] = {"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 bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int dp[maxn][3]; int main(){ int T; cin >> T; while(T--){ cin >> n >> m; dp[1][0] = 0; dp[2][0] = (dp[1][0] + m) % 2; dp[2][1] = 1 - dp[2][0]; dp[3][0] = (dp[2][0] + m) % 3; dp[3][1] = (dp[2][1] + m) % 3; dp[3][2] = 3 - dp[3][1] - dp[3][0]; for(int i = 4; i <= n; ++i) for(int j = 0; j < 3; ++j) dp[i][j] = (dp[i-1][j] + m) % i; for(int i = 0; i < 3; ++i) ++dp[n][i]; cout << dp[n][2] << " " << dp[n][1] << " " << dp[n][0] << endl; } return 0; }