ZOJ-1259

模拟题，直接用两个栈处理就行了，思路要清晰，不难

#include<cstdio>#include<deque>using namespace std;int main(){int N, t;deque<int> A;deque<int> S;while (scanf("%d", &N), N){while (scanf("%d", &t), t){A.clear();S.clear();for (int i = 1; i <= N; i++)A.push_back(i);while (A.front() != t){S.push_back(A.front());A.pop_front();}A.pop_front();bool can = true;for (int i = 2; i <= N; i++){scanf("%d", &t);if (!can)continue;if (!S.empty() && S.back() == t)S.pop_back();else{while (!A.empty() && A.front() != t){S.push_back(A.front());A.pop_front();}if (A.empty())can = false;elseA.pop_front();}}puts(can ? "Yes" : "No");}puts("");}return 0;}