CodeForcesGym 100517B Bubble Sort

Bubble Sort

Time Limit: 2000ms

Memory Limit: 262144KB

This problem will be judged on CodeForcesGym. Original ID: 100517B
64-bit integer IO format: %I64d      Java class name: (Any)

 

解题：我们发现假设当前位选择1,那么发现比1大的并且没有使用的有b个，那么当前为选1，后面就还有$2^{b-1}$种方式，所以贪心的选，只要使得后面的方案数大于当前的k即可，如果少于了怎么办？比如当前是1，然后把当前的值改为2，k减去当前位选1的时候的方案数，类似搞，直到后面的方案的数不少于k

 

代码写的确实烂了点

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 const int maxn = ;
 int c[maxn];
 void add(int i,int val) {
     while(i < maxn) {
         c[i] += val;
         i += i&-i;
     }
 }
 int sum(int i,int ret = ) {
     while(i > ) {
         ret += c[i];
         i -= i&-i;
     }
     return ret;
 }
 bool used[maxn];
 int n,ans[maxn];
 LL k;
 LL check(int x) {
     if(x < ) return ;
     LL ret = ;
     for(int i = ; i < x; ++i){
         if(ret >= k) return ret;
         ret <<= ;
     }
     return ret;
 }
 int main() {
 #define NAME "bubble"
     freopen(NAME".in","r",stdin);
     freopen(NAME".out","w",stdout);
     while(scanf("%d%I64d",&n,&k),n||k) {
         memset(used,false,sizeof used);
         memset(c,,sizeof c);
         for(int i = ; i <= n; ++i) add(i,);
         int cur = ;
         for(int i = ; i < n; ++i) {
             while(used[cur]) ++cur;
             int big = sum(n) - sum(cur);
             if(check(big-) >= k) {
                 ans[i] = cur;
                 used[cur] = true;
                 add(cur,-);
             } else {
                 int t = cur + ;
                 k -= check(big-);
                 while(used[t]) ++t;
                 int xx = sum(n) - sum(t);
                 while(xx && check(xx-) < k) {
                     k -= check(xx-);
                     ++t;
                     while(used[t]) ++t;
                     xx = sum(n) - sum(t);
                 }
                 used[t] = true;
                 ans[i] = t;
                 add(t,-);
             }
         }
         for(int i = ; i <= n; ++i)
             if(!used[i]) {
                 ans[n] = i;
                 break;
             }
         for(int i = ; i <= n; ++i)
             printf("%d%c",ans[i],i==n?'\n':' ');
     }
     return ;
 }
 /*
 1 2 3 4 5
 1 2 3 5 4
 1 2 4 3 5
 1 2 5 3 4
 1 3 2 4 5
 1 3 2 5 4
 1 4 2 3 5
 1 5 2 3 4
 2 1 3 4 5
 2 1 3 5 4
 2 1 4 3 5
 2 1 5 3 4
 3 1 2 4 5
 3 1 2 5 4
 4 1 2 3 5
 5 1 2 3 4
 */