String

Problem Description

You hava a non-empty string which consists of lowercase English letters and may contain at most one '?'. Let's choose non-empty substring G from S (it can be G = S). A substring of a string is a continuous subsequence of the string. if G contains '?' then '?' can be deleted or replaced by one of lowercase english letters. After that if each letter occurs even number of times in G then G is a good substring. Find number of all good substrings.

 

Input

The input consists of an integer T, followed by T lines, each containing a non-empty string. The length of the string doesn't exceed 20000.

[Technical Specification]
1 <= T <= 100

 

Output

For each test case, print a single integer which is the number of good substrings of a given string.

 

Sample Input

3 abc?ca aabbcc aaaaa

 

Sample Output

7 6 6

 

题解：用二进制模拟前缀的奇偶状态，如果一个状态x经过添加了几个字母后回到状态x，那么这几个字母肯定符合good substrings。

感受：又get到了一个新技能。同时深深佩服想出这种方法的大佬。代码有参考网上的博客，不过花了1800多秒，手动实现的map只需要300多秒。

#pragma warning(disable:4996)
#include<map>
#include<string>
#include<cstdio>
#include<bitset>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;

const int maxn = 2e4 + 4;

int T;
char s[maxn];

int main()
{
    scanf("%d", &T); while (T--) {
        map<int, int> p;
        scanf("%s", s);
        int pos = -1, n = strlen(s);
        for (int i = 0; i < n; i++) if (s[i] == '?') { pos = i; break; }

        int ans = 0, x;
        if (pos == -1) {
            x = 0; p[0]++;
            for (int i = 0; i < n; i++) {
                x ^= 1 << (s[i] - 'a');
                ans += p[x];
                p[x]++;
            }
            printf("%d
", ans);
        }
        else {
            x = 0; p[0]++;
            for (int i = 0; i <= pos - 1; i++) {
                x ^= 1 << (s[i] - 'a');
                ans += p[x];
                p[x]++;
            }
            p.clear();

            x = 0; p[0]++;
            for (int i = pos + 1; i < n; i++) {
                x ^= 1 << (s[i] - 'a');
                ans += p[x];
                p[x]++;
            }

            x = 0;
            if (p.count(x)) ans += p[x];
            
            for (int i = 0; i < 26; i++) if (p.count(x ^ (1 << i))) ans += p[x ^ (1 << i)];

            for (int i = pos - 1; i >= 0; i--) {
                x ^= 1 << (s[i] - 'a');
                if (p.count(x)) ans += p[x];
                for (int j = 0; j < 26; j++) if (p.count(x ^ (1 << j))) ans += p[x ^ (1 << j)];
            }
            printf("%d
", ans);
        }
    }
    return 0;
}

网上大佬手动实现的map，值得学习!!!!!!!!!!!!!

#pragma warning(disable:4996)
#include<map>
#include<string>
#include<cstdio>
#include<bitset>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;

const int maxn = 2e4 + 4;

char s[maxn];

struct m_map {
    int tot, k[maxn], va[maxn];
    int head[maxn], next[maxn];
    m_map() { tot = 0; }
    void clear() {
        tot = 0;
        memset(head, -1, sizeof(head));
    }
    int &get(int v) {
        int u = v % maxn;
        for (int i = head[u]; i != -1; i = next[i]) if (k[i] == v) return va[i];
        k[tot] = v; va[tot] = 0;
        next[tot] = head[u];
        head[u] = tot;
        return va[tot++];
    }
    int get2(int v) {
        int u = v % maxn;
        for (int i = head[u]; i != -1; i = next[i]) if (k[i] == v) return va[i];
        return 0;
    }
}mp;

int Find(int p, int n) {
    mp.clear();
    mp.get(0)++;  //初始状态为0，想想"aa"这样的序列
    int ans = 0, x = 0;
    for (int i = p; i < n; i++) {
        x ^= 1 << (s[i] - 'a');
        ans += mp.get2(x);
        mp.get(x)++;
    }
    return ans;
}

int Find2(int p, int n) {
    mp.clear();
    mp.get(0)++;                //同理
    int x = 0, ans = 0;
    for (int i = 0; i < p; i++) {
        x ^= 1 << (s[i] - 'a');
        mp.get(x)++;
    }
    ans += mp.get2(x);
    for (int i = 0; i < 26; i++) ans += mp.get2(x ^ (1 << i));
    for (int i = p + 1; i < n; i++) {
        x ^= 1 << (s[i] - 'a');
        ans += mp.get2(x);
        for (int j = 0; j < 26; j++) ans += mp.get2(x ^ (1 << j));
    }
    return ans;
}

int main()
{
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%s", s);
        int p = -1, n = strlen(s);
        for (int i = 0; i < n; i++) if (s[i] == '?') p = i;
        if (p == -1) printf("%d
", Find(0, n));
        else printf("%d
", Find(0, p) + Find(p + 1, n) + Find2(p, n));
    }
    return 0;
}