A.
题意：给定一个序列，可以把里面的任意几个数加上K的整数倍，变换之后要求把这个序列从小到大排后变为 数列1，2，3，4，5，6
解答：最直观的方法是个匹配，但觉得500*100*100*100可能会超
不过可以发现A可以变为B only when A%K == B%K && A <= K
那么对每个可能的余数开个优先队列扫一遍就可以啦
要被自己蠢死了QAQ
没清空完就break出QAQ
#include <cstdio> #include <iostream> #include <cstring> #include <set> #include <queue> using namespace std; const int N = 105; typedef struct node{ int xi,v; bool operator < (const node &a)const{ if (xi == a.xi) return v < a.v; else return xi > a.xi; } }node; node newnode(int x,int y){ node tmp; tmp.xi = x; tmp.v = y; return tmp; } int n,m,k; int a[N],b[N],tmpa[N],tmpb[N]; priority_queue<node>Q[N]; int main(){ int T; cin >> T; while (T--){ cin >> n >> k; for (int i = 1;i <= n;i++) scanf("%d",&a[i]); for (int i = 1;i <= n;i++) b[i] = i; for (int i = 1;i <= n;i++) Q[a[i]%k].push(newnode(a[i],1)); for (int i = 1;i <= n;i++) Q[b[i]%k].push(newnode(b[i],-1)); int flag = 0; int cnt = 0; for (int t = 0;t < k;t++,cnt = 0){ while(!(Q[t].empty())){ node u = Q[t].top(); Q[t].pop(); cnt += u.v; if (cnt < 0) flag = 1; } if (cnt != 0) flag = 1; } if (flag) printf("Tom\n"); else printf("Jerry\n"); } }
转载于:https://www.cnblogs.com/williamchenwl/p/4839356.html
