Different Integers（牛客多校第一场+莫队做法）

题目链接：https://www.nowcoder.com/acm/contest/139/J

题目：

题意：给你n个数，q次查询，对于每次查询得l，r，求1~l和r~n元素得种类。

莫队思路：1.将元素copy一份到最右边然后对于每次查询得l，r，我们就可以转换成求r，l+n这一个连续区间得元素种类，就将其转换成了一个莫队模板题了（比赛时还不会莫队就随便找了个板子）；

　　　　　2.将移动的两个指针l设为0，r设为n+1，然后进行莫队即可。

做法一代码实现如下：

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cmath>
 #include <cstdlib>
 #include <cstring>
 #include <vector>
 #include <list>
 #include <map>
 #include <stack>
 #include <queue>
 using namespace std;
 #define ll long long
 const int maxn1 = ;
 int cur[maxn1];
 int then[maxn1];
 int ans[maxn1];
 int limit,n,m;
 struct node
 {
     int l,r,id;
 }que[maxn1];
 bool cmp(node x,node y)
 {
     if(x.l/limit == y.l/limit)
         return x.r < y.r;
     return x.l/limit < y.l/limit;
 }
 void init()
 {
     for(int i = ;i <= n;i++) {
         scanf("%d",&cur[i]);
         cur[i+n] = cur[i];
     }
     for(int i = ;i <= m;i++)
     {
         scanf("%d%d",&que[i].l,&que[i].r);
         int t = que[i].r;
         que[i].r = que[i].l + n;
         que[i].l = t;
         que[i].id = i;
     }
     limit = (int)(sqrt( * n)+0.5);
     memset(then,,sizeof(then));
     memset(ans, , sizeof(ans));
     sort(que+,que++m,cmp);
 }
 void solve()
 {
     int L,R,ans1;
     L = R = ;
     ans1 = ;
     for(int i = ;i <= m;i++)
     {
         while(que[i].l > L)
         {
             then[cur[L]]--;
             if(then[cur[L]] == )
                 ans1--;
             L++;
         }
         while(que[i].r < R)
         {
             then[cur[R]]--;
             if(then[cur[R]] == )
                 ans1--;
             R--;
         }
         while(que[i].l < L)
         {
             L--;
             then[cur[L]]++;
             if(then[cur[L]] == )
                 ans1++;
         }
         while(que[i].r > R)
         {
             R++;
             then[cur[R]]++;
             if(then[cur[R]] == )
                 ans1++;
         }
         ans[que[i].id] = ans1;
     }
     for(int i = ;i <= m;i++)
         printf("%d\n",ans[i]);
 }
 int main()
 {
     while(~scanf("%d%d", &n, &m)) {
         init();
         solve();
     }
     return ;
 }

做法二代码实现如下：

 #include <set>
 #include <map>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <bitset>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 typedef long long ll;
 typedef unsigned long long ull;

 #define bug printf("*********\n");
 #define FIN freopen("in.txt", "r", stdin);
 #define debug(x) cout<<"["<<x<<"]" <<endl;
 #define IO ios::sync_with_stdio(false),cin.tie(0);

 const double eps = 1e-;
 const int mod = 1e9 + ;
 const int maxn = 1e5 + ;
 const double pi = acos(-);
 const int inf = 0x3f3f3f3f;
 const ll INF = 0x3f3f3f3f3f3f3f3f;

 inline int read() {//读入挂
     int ret = , c, f = ;
     for(c = getchar(); !(isdigit(c) || c == '-'); c = getchar());
     if(c == '-') f = -, c = getchar();
     for(; isdigit(c); c = getchar()) ret = ret *  + c - '';
     if(f < ) ret = -ret;
     return ret;
 }

 int n, q, sum, block;
 int a[maxn], cnt[maxn], pos[maxn];

 struct node {
     int l, r, ans, id;
 }ask[maxn];

 bool cmp(const node& x, const node& y) {
     return pos[x.l] == pos[y.l] ? x.r < y.r : x.l < y.l;
 }

 void add(int x) {
     cnt[x]++;
     if(cnt[x] == ) sum++;
 }

 void del(int x) {
     if(cnt[x] == ) sum--;
     cnt[x]--;
 }

 int main() {
     //FIN;
     while(~scanf("%d%d", &n, &q)) {
         for(int i = ; i <= n; i++) {
             cnt[i] = ;
         }
         sum = ;
         block = sqrt( * n);
         for(int i = ; i <= n; i++) {
             a[i] = read();
             pos[i] = (i - ) / block;
         }
         for(int i = ; i <= q; i++) {
             ask[i].l = read();
             ask[i].r = read();
             ask[i].id = i;
         }
         sort(ask + , ask + q + , cmp);
         for(int i = , l = , r = n + ; i <= q; i++) {
             while(r < ask[i].r) del(a[r++]);
             while(r > ask[i].r) add(a[--r]);
             while(l < ask[i].l) add(a[++l]);
             while(l > ask[i].l) del(a[l--]);
             ask[ask[i].id].ans = sum;
         }
         for(int i = ; i <= q; i++)
             printf("%d\n", ask[i].ans);
     }
     return ;
 }