The 2019 Asia Nanchang First Round Online Programming Contest E. Magic Master
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题目链接:https://nanti.jisuanke.com/t/41352
题目意思还是好理解的,看过的人不多,感觉是被通过量吓到了。其实就是个水题,反向模拟就好了,
用队列模拟,反向模拟,它要放m张卡到后面,那我就放m张卡到前面,一开始队列只有1张卡,慢慢加到n张卡,
先加大的卡,再一直到1的卡。对了,可能会出现只有5张卡,m却是6,7,8或9,10,那么为了减少不必要的模拟,
用mod来减少,因为有些模拟会让卡和之前比较,算是原封不动。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <queue> 6 #include <stack> 7 #include <string> 8 #include <map> 9 #include <cmath> 10 #include <iomanip> 11 using namespace std; 12 13 typedef long long LL; 14 #define inf 1e9 15 #define rep(i,j,k) for(int i = (j); i <= (k); ++i) 16 #define rep__(i,j,k) for(int i = (j); i < (k); ++i) 17 #define per(i,j,k) for(int i = (j); i >= (k); --i) 18 #define per__(i,j,k) for(int i = (j); i > (k); --i) 19 20 const int N = 40000010; 21 int arr[N]; 22 int mod,n,q; 23 queue<int> que; 24 25 void work() 26 27 int tmp; 28 que.push(n);//压入最大的卡 29 int num = 1; 30 int end = n - 1;//最后1的卡直接是取走的,所以那一轮不需要模拟了 31 while(num <= end) 32 tmp = mod % num; 33 34 rep(o,1,tmp) 35 que.push(que.front()); 36 que.pop(); 37 38 que.push(n-num);//压入前一张卡 39 ++num; 40 41 // que.push(1); 42 43 44 int main() 45 46 int T; 47 scanf("%d",&T); 48 49 while(T--) 50 scanf("%d%d",&n,&mod); 51 52 rep(i,1,n) 53 arr[i] = i;//1-n的卡 54 55 scanf("%d",&q); 56 57 work(); 58 59 per(i,n,1)//取出卡 60 arr[i] = que.front(); 61 que.pop(); 62 63 64 int o; 65 rep(i,1,q) 66 scanf("%d",&o); 67 printf("%d\n",arr[o]); 68 69 70 // rep(i,1,n) printf("%d ",arr[i]); 71 72 73 getchar(); getchar(); 74 return 0; 75
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