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ORZ,这道题想复杂了,原来直接暴力就可以了复杂度为 n * n * logn
#include<cstdio> #include<set> #include<algorithm> using namespace std; typedef long long LL; const int maxn = 10005; set<LL>s; set<LL>::iterator it1,it2; int n,m; int num[maxn] = {0}; LL arr[maxn]; int main(){ scanf("%d%d",&n,&m); for(int i = 0; i < n; i++) scanf("%I64d",&arr[i]); for(int i = 0; i < n; i++){ s.clear(); s.insert(arr[i]); num[1] ++; for(int j = i + 1; j < n; j++){ if(!s.count(arr[j])) s.insert(arr[j]); else break; it1 = s.begin(); it2 = s.end(); it2 --; LL e1 = *it1,e2 = *it2; if(e2 - e1 == s.size() - 1){ //printf("%I64d %I64d\n",e1,e2); num[s.size()] ++; } } } printf("Case #1:\n"); while(m--){ int k; scanf("%d",&k); printf("%d\n",num[k]); } return 0; }
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原文地址:http://blog.csdn.net/u013451221/article/details/46346237