洛谷P1494 小Z的袜子

题意：在[l, r]之中任选两个数，求它们相同的概率。

解：

莫队入门。

概率这个很好搞，就是cnt * (cnt - 1) / 2。

然后发现每次挪指针的时候，某一个cnt会+1或-1。这时候差值就是2 * cnt_small。

 #include <cstdio>
 #include <algorithm>
 #include <cmath>
 
 typedef long long LL;
 const int N = ;
 
 int a[N], bin[N], fr[N];
 LL ans;
 
 struct ASK {
     int id, l, r;
     LL ans;
     inline bool operator <(const ASK &w) const {
         if(fr[l] != fr[w.l]) {
             return fr[l] < fr[w.l];
         }
         return r < w.r;
     }
 }ask[N];
 
 inline bool cmp(const ASK &a, const ASK &b) {
     return a.id < b.id;
 }
 
 inline LL gcd(LL a, LL b) {
     if(!b) {
         return a;
     }
     return gcd(b, a % b);
 }
 
 inline void add(int x) {
     ans += bin[a[x]] << ;
     bin[a[x]]++;
     return;
 }
 
 inline void del(int x) {
     bin[a[x]]--;
     ans -= bin[a[x]] << ;
 }
 
 int main() {
     int n, m;
     scanf("%d%d", &n, &m);
     int T = sqrt(n);
     for(int i = ; i <= n; i++) {
         scanf("%d", &a[i]);
         fr[i] = (i - ) / T + ;
     }
     for(int i = ; i <= m; i++) {
         ask[i].id = i;
         scanf("%d%d", &ask[i].l, &ask[i].r);
     }
     std::sort(ask + , ask + m + );
 
     int L = , R = ;
     bin[a[]]++;
     for(int i = ; i <= m; i++) {
         if(ask[i].l == ask[i].r) {
             continue;
         }
         while(ask[i].l < L) {
             add(--L);
         }
         while(R < ask[i].r) {
             add(++R);
         }
         while(L < ask[i].l) {
             del(L++);
         }
         while(ask[i].r < R) {
             del(R--);
         }
         ask[i].ans = ans;
     }
 
     std::sort(ask + , ask + m + , cmp);
     for(int i = ; i <= m; i++) {
         if(!ask[i].ans || ask[i].l == ask[i].r) {
             puts("0/1");
             continue;
         }
         int g = gcd(ask[i].ans, 1ll * (ask[i].r - ask[i].l + ) * (ask[i].r - ask[i].l));
         printf("%lld/%lld\n", ask[i].ans / g, 1ll * (ask[i].r - ask[i].l + ) * (ask[i].r - ask[i].l) / g);
     }
     return ;
 }

AC代码