模拟_CH0802_占卜DIY

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思路分析:

    本题直接模拟即可, 为方便, 本人使用deque表示每堆牌, AC代码如下:

//CH0802_占卜DIY
#include <iostream>
#include <cstdio>
#include <deque>
#include <algorithm>
using namespace std;
pair<deque<int>, int> heap[14];//first[0...3]堆中牌(从上到下), second:本堆未掀开牌数 
int kcnt;//当前尚未掀开的K的个数 
void play(){
	if(!kcnt) return;
	int q = heap[13].first[0]; heap[13].first.pop_front();
	while(true){
		if(q == 13){
			--kcnt, play(); 
			return;
		}
		heap[q].first.push_front(q), --heap[q].second;
		int p = heap[q].first[4];
		heap[q].first.pop_back();
		if((q = p) == 13){
			--kcnt, play(); 
			return;
		}
	}
}
int main(){
	char ch;
	for(int i = 1, tmp; i <= 13; ++i) 
		for(int j = 1; j <= 4; ++j) 
			cin >> ch
			, tmp = ch == 'A'? 1: ch == 'J'? 11: ch == 'Q'? 12: ch == 'K'? 13: ch == '0'? 10: ch - '0'
			, heap[i].first.push_back(tmp), heap[i].second = 4;
	kcnt = 4, play();
	int res = 0; for(int i = 1; i <= 12; ++i) if(!heap[i].second) ++res;
	cout << res << endl;
	return 0;
}