2017 Multi-University Training Contest 6 solutions 1003 Inversion
2017-08-11 19:45
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需要统计的区间最大值次数是OnlognO(nlog(n))级别的。
时间复杂度 OnlognO(nlog(n))。
优秀的多个OlognO(log(n))的做法也是可以卡过的。
这里介绍一下优雅的暴力做法。将AA数组按值从大到小排序,对于每个下标ii暴力找到最大的不被ii整除的数。
时间复杂度 OnlognO(nlog(n))。
第一次Compilation Error,用Xcode写的注释mingw无法识别见鬼了,居然罚了20分钟(不可理喻)
把注释去掉即可ac,
现在外面交不去注释可直接ac
气!!!
时间复杂度 OnlognO(nlog(n))。
优秀的多个OlognO(log(n))的做法也是可以卡过的。
这里介绍一下优雅的暴力做法。将AA数组按值从大到小排序,对于每个下标ii暴力找到最大的不被ii整除的数。
时间复杂度 OnlognO(nlog(n))。
// // main.cpp // Inversion // // Created by wenhan on 2017/8/10. // Copyright �� 2017年 wenhan. All rights reserved. // #include <iostream> #include <cstdio> #include <algorithm> using namespace std; struct me{ int s; int p; }me[100000+5]; int cmp(struct me a,struct me b) { if(a.s==b.s) return a.p<b.p; return a.s<b.s; } int main() { int t; scanf("%d",&t); while (t--) { int n; scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%d",&me[i].s); me[i].p=i; } sort(me+1, me+1+n,cmp); for(int i=2;i<=n;i++) { for(int j=n;j>=1;j--) { if(me[j].p%i!=0) { if(i==2) printf("%d",me[j].s); else printf(" %d",me[j].s); break; } } } printf("\n"); } // insert code here... //std::cout << "Hello, World!\n"; return 0; }
第一次Compilation Error,用Xcode写的注释mingw无法识别见鬼了,居然罚了20分钟(不可理喻)
把注释去掉即可ac,
现在外面交不去注释可直接ac
气!!!
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