HDU 5056 - Boring count

题意

统计一个字符串中子串每个字符最多重复k次的子串。

思路

用pos记录位置，当碰到cnt>k的时候就往前移

代码

  
  
  
  

   
   
   
   
#include <cstdio>
   
   
   
   
#include <stack>
   
   
   
   
#include <set>
   
   
   
   
#include <iostream>
   
   
   
   
#include <string>
   
   
   
   
#include <vector>
   
   
   
   
#include <queue>
   
   
   
   
#include <functional>
   
   
   
   
#include <cstring>
   
   
   
   
#include <algorithm>
   
   
   
   
#include <cctype>
   
   
   
   
#include <ctime>
   
   
   
   
#include <cstdlib>
   
   
   
   
#include <fstream>
   
   
   
   
#include <string>
   
   
   
   
#include <sstream>
   
   
   
   
#include <map>
   
   
   
   
#include <cmath>
   
   
   
   
#define LL long long
   
   
   
   
#define lowbit(x) ((x) & (-x))
   
   
   
   
#define MP(a, b) make_pair(a, b)
   
   
   
   
#define MS(arr, num) memset(arr, num, sizeof(arr))
   
   
   
   
#define PB push_back
   
   
   
   
#define F first
   
   
   
   
#define S second
   
   
   
   
#define ROP freopen("input.txt", "r", stdin);
   
   
   
   
#define BitCount(x) __builtin_popcount(x)
   
   
   
   
const double PI = acos(-1.0);
   
   
   
   
const int INF = 0x3f3f3f3f;
   
   
   
   
using namespace std;
   
   
   
   
const int MAXN = 1e5 + 100;
   
   
   
   
const int MOD = 20071027;
   
   
   
   

   
   
   
   
typedef pair<int, int> pii;
   
   
   
   
typedef vector<int>::iterator viti;
   
   
   
   
typedef vector<pii>::iterator vitii;
   
   
   
   

   
   
   
   
char str[MAXN];
   
   
   
   
int cnt[30];
   
   
   
   

   
   
   
   
int main()
   
   
   
   
{
   
   
   
   
 //ROP;
   
   
   
   
 int n, i, j, last, T, k;
   
   
   
   
 scanf("%d", &T);
   
   
   
   
 while (T--)
   
   
   
   
 {
   
   
   
   
 MS(cnt, 0);
   
   
   
   
 int pos = 1;
   
   
   
   
 scanf("%s", str + 1);
   
   
   
   
 scanf("%d", &k);
   
   
   
   
 int len = strlen(str + 1);
   
   
   
   
 LL ans = 0;
   
   
   
   
 for (i = 1; i <= len; i++)
   
   
   
   
 {
   
   
   
   
 cnt[str[i] - 'a']++;
   
   
   
   
 while (cnt[str[i] - 'a'] > k)
   
   
   
   
 {
   
   
   
   
 cnt[str[pos] - 'a']--;
   
   
   
   
 pos++;
   
   
   
   
 }
   
   
   
   
 ans += i - pos + 1;
   
   
   
   
 }
   
   
   
   
 cout << ans << endl;
   
   
   
   
 }
   
   
   
   
 return 0;
   
   
   
   
}