Diverse Substrings

合集 - 刷题集(97)

1.妖梦拼木棒2023-05-232.[USACO1.2]回文平方数 Palindromic Squares2023-05-243.[COCI2012-2013#1] LJUBOMORA2023-05-254.火柴棒等式2023-05-255.Kieszonkowe2023-05-256.[COCI2014-2015#5] TRAKTOR2023-05-267.硬币翻转2023-05-268.[蓝桥杯 2020 省 AB1] 走方格2023-05-269.P2097 资料分发2023-05-2610.[AHOI2005]约数研究2023-05-2611.[NOIP2004 普及组] 火星人2023-05-2612.我焯,原神(这真是个题)2023-05-2613.滑雪 OJ36512023-05-2414.CV的人有福啦!YTU贪心训练2(部分注释)2023-05-2715.砝码称重2023-05-2716.警钟长鸣-----找零钱2023-05-2817.[USACO09OCT]Barn Echoes G2023-05-2818.【CSGRound2】光骓者的荣耀2023-05-2919.初见蓝桥杯2023-05-3020.[蓝桥杯 2022 省 B] 扫雷2023-05-3121.开餐馆2023-05-3122.P9241 [蓝桥杯 2023 省 B] 飞机降落2023-06-0123.P7083 [NWRRC2013] Energy Tycoon2023-06-0324.高精度2023-06-0425.P7954 [COCI2014-2015#6] PAPRIKA2023-06-0426.(输出路径搜索)[USACO06OCT] Cows on Skates G2023-06-0427.畜栏保留问题2023-06-0428.(贪心+搜索+剪枝)P8801 [蓝桥杯 2022 国 B] 最大数字2023-06-0529.蓝桥杯 入门----普及-2023-06-0630.[NOIP2016 提高组] 玩具谜题2023-06-0631.[NOIP2002 提高组] 字串变换(无向图查重)2023-06-0632.[NOIP2000 提高组] 单词接龙2023-06-0733.POJ 硬币2023-06-0734.动态规划学习 1(最长上升子序列问题)2023-06-0735.动态规划学习 2(背包问题及其变形)2023-06-0836. 切割回文2023-06-1037.[NOIP2001 提高组] 数的划分(剪枝)2023-06-1038.字符串转换数字,sscanf和sprintf大法2023-06-1039.[CTSC1997] 选课(树状DP)2023-06-1040. k 倍区间(同余定理,组合数)2023-06-1141.[蓝桥杯 2018 省 AB] 全球变暖2023-06-1142.马的遍历(对bfs的更深一层探讨)2023-06-1143.找零钱2023-06-1144.Pasha Maximizes2023-06-1245.Newspaper Headline2023-06-1246.Balanced Ternary String2023-06-1347.(数论)素数,质数2023-06-1348.(数论) 约数2023-06-1349.Helping the Nature2023-06-1450.(博弈论)Even Number Addicts2023-06-1451.Present2023-06-1552.Mr. Kitayuta's Colorful Graph(二维并查集,弱化版)2023-06-1653.Edgy Trees(dfs,并查集,快速幂,树形结构,红黑树)2023-06-1654.并查集2023-05-2455.再谈 前缀和,差分,离散化2023-06-1656.Best Cow Fences(前缀和+特殊二分)2023-06-1757.二分2023-06-1758.同类型,类背包动态规划,选地dp2023-06-1759.Primes on Interval(欧拉筛+二分+滑动窗口)2023-06-1860.字符串(KMP , Trie树 , STL)2023-06-1861.Codeforces1 #879 div.22023-06-1962.Add Modulo 10(数论,思维,数学,规律)2023-06-2063.期末考试YTU4035: Shmily(数学,等差数列)2023-06-2064.[POI2006] OKR-Periods of Words2023-06-2165.Getting Zero(Bfs)2023-06-2166.快速幂2023-06-2167.Dreaming of Freedom(数论,贪心)2023-06-2368.Scoring Subsequences2023-06-2469.Dolce Vita2023-06-2970.Cake Assembly Line2023-06-2971.Constructive Problem2023-06-3072.Forever Winter2023-06-3073.Maximum Sum(前缀和)2023-07-1074.Contrast Value(数学规律)2023-07-1075.Lamps(STL+双端队列)2023-07-1076.Long Long2023-07-1177.Sum in Binary Tree2023-07-1178.Apple Tree(树状搜索,树形DP)2023-07-1179.Strong Password(贪心思想)2023-07-1280.Come Together2023-07-1281.Notepad2023-07-1382.Hamiltonian Wall2023-07-1383.7.17作业2023-07-1684.逛画展(双指针)2023-07-1985.#8852023-07-2086.树状数组2023-07-2087.Subsequence Addition2023-07-2288.7.22作业2023-07-22

89.Diverse Substrings2023-07-24

90.Tracking Segments(二分,区间前缀)2023-07-2791.最大食物链计数2023-07-2792.线段树12023-07-2293.[TJOI2007] 线段(状态机)2023-07-2994.#8782023-07-3095.测试12023-07-3096.李白打酒2023-07-3097.砍竹子2023-07-30

收起

# Diverse Substrings

## 题面翻译

定义一个数字串是**多变的**当且仅当其中所有数字的重复次数均不超过其中不同数字的种类数。

给定一个由 $0$ 到 $9$ 组成的长为 $n$ 的数字串 $s$，求其不同的**多变的**子串 $s_{[l,r]}$ 的个数。

## 题目描述

A non-empty digit string is diverse if the number of occurrences of each character in it doesn't exceed the number of distinct characters in it.

For example:

- string "7" is diverse because 7 appears in it $1$ time and the number of distinct characters in it is $1$ ;
- string "77" is not diverse because 7 appears in it $2$ times and the number of distinct characters in it is $1$ ;
- string "1010" is diverse because both 0 and 1 appear in it $2$ times and the number of distinct characters in it is $2$ ;
- string "6668" is not diverse because 6 appears in it $3$ times and the number of distinct characters in it is $2$ .

You are given a string $s$ of length $n$ , consisting of only digits $0$ to $9$ . Find how many of its $\frac{n(n+1)}{2}$ substrings are diverse.

A string $a$ is a substring of a string $b$ if $a$ can be obtained from $b$ by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

Note that if the same diverse string appears in $s$ multiple times, each occurrence should be counted independently. For example, there are two diverse substrings in "77" both equal to "7", so the answer for the string "77" is $2$ .

## 输入格式

Each test contains multiple test cases. The first line contains a single integer $t$ ( $1\le t\le {10}^4$ ) — the number of test cases.

The first line of each test case contains a single integer $n$ ( $1\le n\le {10}^5$ ) — the length of the string $s$ .

The second line of each test case contains a string $s$ of length $n$ . It is guaranteed that all characters of $s$ are digits from $0$ to $9$ .

It is guaranteed that the sum of $n$ over all test cases does not exceed ${10}^5$ .

## 输出格式

For each test case print one integer — the number of diverse substrings of the given string $s$ .

## 样例 #1

### 样例输入 #1

```
7
1
7
2
77
4
1010
5
01100
6
399996
5
23456
18
789987887987998798
```

### 样例输出 #1

```
1
2
10
12
10
15
106
```

## 提示

In the first test case, the diverse substring is "7".

In the second test case, the only diverse substring is "7", which appears twice, so the answer is $2$ .

In the third test case, the diverse substrings are "0" ( $2$ times), "01", "010", "1" ( $2$ times), "10" ( $2$ times), "101" and "1010".

In the fourth test case, the diverse substrings are "0" ( $3$ times), "01", "011", "0110", "1" ( $2$ times), "10", "100", "110" and "1100".

In the fifth test case, the diverse substrings are "3", "39", "399", "6", "9" ( $4$ times), "96" and "996".

In the sixth test case, all $15$ non-empty substrings of "23456" are diverse.

//由于题目要求,可以得到,每个数都有一个最大的限制范围,考虑最差情况,如果每个数都不同
//那么只需要100位即可,因为,每个数的大小最多不能超过9,如果到100,那么一定会有一个数会成为第11个数
//所以从i位开始,只需要枚举之后的最多i+100位就可以了
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N=1e6+10,mod=1e9+7;
string s;
int n,t,a[N],res,num,ans,m,f[11];
signed main()
{
    std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>t;
    while(t--){
        res=0;
        cin>>n>>s;
        for(int i=0;i<min(n,(int)(i+101));i++){
            num=0;memset(f,0,sizeof f);
            for(int j=i;j<min(n,(int)(i+101));j++){
                if(f[s[j]-'0']==0) num++;
                f[s[j]-'0']++;
                bool a=false;
                for(int k=0;k<10;k++) if(f[k]>num) a=true;
                if(!a) res++;
            }
        }
        cout<<res<<endl;
    }
    return 0;
}

 

__EOF__

本文作者：Sakurajimamai
本文链接：https://www.cnblogs.com/o-Sakurajimamai-o/p/17576775.html
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