• HDU6345 子串查询


    莫队

    直接莫队就好了。。统计每个字母出现的次数= =

    #include <bits/stdc++.h>
    #define INF 0x3f3f3f3f
    #define full(a, b) memset(a, b, sizeof a)
    #define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
    using namespace std;
    typedef long long ll;
    inline int lowbit(int x){ return x & (-x); }
    inline int read(){
        int ret = 0, w = 0; char ch = 0;
        while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
        while(isdigit(ch)) ret = (ret << 3) + (ret << 1) + (ch ^ 48), ch = getchar();
        return w ? -ret : ret;
    }
    inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }
    inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
    template <typename T>
    inline T max(T x, T y, T z){ return max(max(x, y), z); }
    template <typename T>
    inline T min(T x, T y, T z){ return min(min(x, y), z); }
    template <typename A, typename B, typename C>
    inline A fpow(A x, B p, C lyd){
        A ans = 1;
        for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
        return ans;
    }
    const int N = 200005;
    int _, n, q, t, cnt[30], res[N];
    char s[N];
    struct Query{
        int l, r, id, block;
        bool operator < (const Query &rhs) const {
            return (block ^ rhs.block) ? l < rhs.l : (block & 1) ? r < rhs.r : r > rhs.r;
        }
    }query[N];
    int main(){
    
        int cs = 0;
        for(_ = read(); _; _ --){
            full(cnt, 0), full(res, 0);
            n = read(), q = read();
            scanf("%s", s + 1);
            t = sqrt(n);
            for(int i = 1; i <= q; i ++){
                query[i].l = read(), query[i].r = read();
                query[i].id = i, query[i].block = (query[i].l - 1) / t + 1;
            }
            sort(query + 1, query + q + 1);
            int l = 1, r = 0;
            for(int i = 1; i <= q; i ++){
                int curL = query[i].l, curR = query[i].r;
                while(l < curL) cnt[s[l++] - 'A'] --;
                while(r < curR) cnt[s[++r] - 'A'] ++;
                while(l > curL) cnt[s[--l] - 'A'] ++;
                while(r > curR) cnt[s[r--] - 'A'] --;
                for(int j = 0; j < 26; j ++){
                    if(cnt[j] != 0){
                        res[query[i].id] = cnt[j];
                        break;
                    }
                }
            }
            printf("Case #%d:
    ", ++cs);
            for(int i = 1; i <= q; i ++) printf("%d
    ", res[i]);
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/onionQAQ/p/11192801.html
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