http://www.gdutcode.sinaapp.com/problem.php?cid=1049&pid=3
dp[i][state]表示处理了前i个，然后当前状态是state的时候的最小休息天数。
比如用1代表休息，2是训练，3是运动。
如果当前这天允许的状态只是训练。
那么dp[i][3] = inf（表示不可能）
休息是无论何时都可能的了。从dp[i - 1][1--3]个状态转过来， ＋ １即可
那个求导是-24
#include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <algorithm> #include <assert.h> #define IOS ios::sync_with_stdio(false) using namespace std; #define inf (0x3f3f3f3f) typedef long long int LL; #include <iostream> #include <sstream> #include <vector> #include <set> #include <map> #include <queue> #include <string> const int maxn = 100 + 20; int dp[maxn][4]; void work() { int n; scanf("%d", &n); memset(dp, 0, sizeof dp); for (int i = 1; i <= n; ++i) { int x; scanf("%d", &x); if (x == 0) { dp[i][1] = min(dp[i - 1][1], min(dp[i - 1][2], dp[i - 1][3])) + 1; dp[i][2] = dp[i][3] = inf; } else if (x == 1) { dp[i][2] = dp[i - 1][1]; dp[i][2] = min(dp[i][2], dp[i - 1][3]); dp[i][1] = min(dp[i - 1][1], min(dp[i - 1][2], dp[i - 1][3])) + 1; dp[i][3] = inf; } else if (x == 2) { dp[i][2] = inf; dp[i][3] = min(dp[i - 1][1], dp[i - 1][2]); dp[i][1] = min(dp[i - 1][1], min(dp[i - 1][2], dp[i - 1][3])) + 1; } else if (x == 3) { dp[i][1] = min(dp[i - 1][1], min(dp[i - 1][2], dp[i - 1][3])) + 1; dp[i][2] = min(dp[i - 1][1], dp[i - 1][3]); dp[i][3] = min(dp[i - 1][1], dp[i - 1][2]); } } int ans = inf; for (int i = 1; i <= 3; ++i) { ans = min(ans, dp[n][i]); } cout << ans * (-24) << endl; } int main() { #ifdef local freopen("data.txt", "r", stdin); // freopen("data.txt", "w", stdout); #endif int t; scanf("%d", &t); while (t--) work(); return 0; }