[洛谷P3709]大爷的字符串题

题目传送门

不用管它随机什么的，就用贪心的思想去想，

会发现这道题的实质是：求查询区间众数出现次数。

莫队即可解决。

注意字符集1e9，要离散化处理。

 #include <bits/stdc++.h>

 using namespace std;

 #define re register
 #define rep(i, a, b) for (re int i = a; i <= b; ++i)
 #define repd(i, a, b) for (re int i = a; i >= b; --i)
 #define maxx(a, b) a = max(a, b);
 #define minn(a, b) a = min(a, b);
 #define LL long long
 #define INF (1 << 30)

 inline int read() {
     int w = , f = ; char c = getchar();
     while (!isdigit(c)) f = c == '-' ? - : f, c = getchar();
     while (isdigit(c)) w = (w << ) + (w << ) + (c ^ ''), c = getchar();
     return w * f;
 }

 const int maxn = 2e5 + ;

 struct Query {
     int l, r, id, pos;
 } q[maxn];
 bool cmp(Query a, Query b) { return a.pos < b.pos || a.pos == b.pos && a.r < b.r; }

 struct Value {
     int v, id;
 } v[maxn];
 bool cmpv(Value a, Value b) { return a.v < b.v; }

 int a[maxn], b[maxn], l[maxn], cnt[maxn], n, m, size;

 int main() {
     n = read(), m = read();
     size = sqrt(n);
     rep(i, , n) v[i].v = read(), v[i].id = i;
     sort(v+, v+n+, cmpv);
     a[v[].id] = ;
     rep(i, , n) a[v[i].id] = a[v[i-].id] + (v[i].v == v[i-].v ?  : );

     rep(i, , m) q[i].l = read(), q[i].r = read(), q[i].id = i, q[i].pos = (q[i].l + size - ) / size;
     sort(q+, q+m+, cmp);
     int pl = , pr = , ans = ; l[] = maxn;
     rep(i, , m) {
         while (pl < q[i].l) l[cnt[a[pl]]]--, l[cnt[a[pl]]-]++, cnt[a[pl++]]--;
         while (pl > q[i].l) cnt[a[--pl]]++, l[cnt[a[pl]]]++, l[cnt[a[pl]]-]--, maxx(ans, cnt[a[pl]]);
         while (pr < q[i].r) cnt[a[++pr]]++, l[cnt[a[pr]]]++, l[cnt[a[pr]]-]--, maxx(ans, cnt[a[pr]]);
         while (pr > q[i].r) l[cnt[a[pr]]]--, l[cnt[a[pr]]-]++, cnt[a[pr--]]--;
         while (!l[ans]) ans--;
         b[q[i].id] = ans;
     }
     rep(i, , m) printf("%d\n", -b[i]);
     return ;
 }