CodeForcesGym 100524J Jingles of a String

Jingles of a String

Time Limit: 2000ms

Memory Limit: 524288KB

This problem will be judged on CodeForcesGym. Original ID: 100524J
64-bit integer IO format: %I64d      Java class name: (Any)

 

 

解题：题目很厉害啊！借鉴某大牛的思路。

 

由于都是小写字母，所以对任意一个区间，最多只有26个，用二进制表示该区间出现过的字母

 

然后就是固定一个左边界，可以算出左边界到字符串末这区间中1个个数，那么枚举这些1的个数，进行二分查找出现这么多个1的区间的最大右边界。

 

RMQ st可以这么用，也是涨姿势了

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 const int maxn = ;
 int st[maxn][];
 char str[maxn];
 int query(int L,int R) {
     int lg =  - __builtin_clz(R - L + );
     return st[L][lg]|st[R - (<<lg) + ][lg];
 }
 unordered_map<int,int>ump;
 int main() {
 #define NAME "jingles"
     freopen(NAME".in","r",stdin);
     freopen(NAME".out","w",stdout);
     int kase;
     scanf("%d",&kase);
     while(kase--) {
         scanf("%s",str);
         int len = strlen(str);
         for(int i = ; i < len; ++i)
             st[i][] = <<(str[i] - 'a');
         for(int j = ; (<<j) <= len; ++j) {
             for(int i = ; i + (<<j) <= len; ++i)
                 st[i][j] = st[i][j-]|st[i+(<<(j-))][j-];
         }
         ump.clear();
         for(int i = ; i < len; ++i) {
             int x = query(i,len - );
             ump[x] = max(ump[x],len - i);
             int cnt = __builtin_popcount(x);
             for(int j = ; j < cnt; ++j) {
                 int low = i,high = len-,ret;
                 while(low <= high) {
                     int mid = (low + high)>>;
                     int y = __builtin_popcount(query(i,mid));
                     if(y <= j) {
                         low = mid + ;
                         ret = mid;
                     } else high = mid - ;
                 }
                 ump[query(i,ret)] = max(ump[query(i,ret)],ret - i + );
             }
         }
         LL ans = ;
         for(auto &it:ump)
             ans += __builtin_popcount(it.first)*(LL)it.second;
         printf("%d %I64d\n",ump.size(),ans);
     }
     return ;
 }