hdu 3333 Turing Tree

题目链接

给n个数， m个询问， 每次询问输出区间内的数的和， 相同的数只计算一次。

数组里的数是>-1e9 <1e9， 可以把它离散以后用莫队搞...

 #include <iostream>
 #include <vector>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include <map>
 #include <set>
 #include <string>
 #include <queue>
 #include <stack>
 #include <bitset>
 using namespace std;
 #define pb(x) push_back(x)
 #define ll long long
 #define mk(x, y) make_pair(x, y)
 #define lson l, m, rt<<1
 #define mem(a) memset(a, 0, sizeof(a))
 #define rson m+1, r, rt<<1|1
 #define mem1(a) memset(a, -1, sizeof(a))
 #define mem2(a) memset(a, 0x3f, sizeof(a))
 #define rep(i, n, a) for(int i = a; i<n; i++)
 #define fi first
 #define se second
 typedef pair<int, int> pll;
 const double PI = acos(-1.0);
 const double eps = 1e-;
 const int mod = 1e9+;
 const int inf = ;
 const int dir[][] = { {-, }, {, }, {, -}, {, } };
 const int maxn = 3e5+;
 int a[maxn], b[maxn], c[maxn];
 ll ans[];
 struct node
 {
     int l, r, id, block;
     bool operator < (node a)const
     {
         if(block == a.block)
             return r<a.r;
         return block<a.block;
     }
 }q[];
 int main()
 {
     int t, n, m;
     cin>>t;
     while(t--) {
         scanf("%d", &n);
         mem(c);
         for(int i = ; i<=n; i++) {
             scanf("%d", &a[i]);
             b[i-] = a[i];
         }
         sort(b, b+n);
         int cnt = unique(b, b+n)-b;
         for(int i = ; i<=n; i++) {
             a[i] = lower_bound(b, b+cnt, a[i])-b+;
         }
         int BLOCK = max(1.0, sqrt(cnt*1.0));
         cin>>m;
         for(int i = ; i<m; i++) {
             scanf("%d%d", &q[i].l, &q[i].r);
             q[i].block = q[i].l/BLOCK;
             q[i].id = i;
         }
         sort(q, q+m);
         ll sum = ;
         for(int i = q[].l; i<=q[].r; i++) {
             if(!c[a[i]]) {
                 sum += b[a[i]-];
             }
             c[a[i]]++;
         }
         ans[q[].id] = sum;
         for(int i = ; i<m; i++) {
             if(q[i].l<q[i-].l) {
                 for(int j = q[i-].l-; j>=q[i].l; j--) {
                     if(!c[a[j]]) {
                         sum += b[a[j]-];
                     }
                     c[a[j]]++;
                 }
             } else {
                 for(int j = q[i-].l; j<q[i].l; j++) {
                     if(c[a[j]]==) {
                         sum -= b[a[j]-];
                     }
                     c[a[j]]--;
                 }
             }
             if(q[i].r>q[i-].r) {
                 for(int j = q[i-].r+; j<=q[i].r; j++) {
                     if(!c[a[j]]) {
                         sum += b[a[j]-];
                     }
                     c[a[j]]++;
                 }
             } else {
                 for(int j = q[i-].r; j>q[i].r; j--) {
                     if(c[a[j]] == ) {
                         sum -= b[a[j]-];
                     }
                     c[a[j]]--;
                 }
             }
             ans[q[i].id] = sum;
         }
         for(int i = ; i<m; i++)
             printf("%I64d\n", ans[i]);
     }
     return ;
 }