Given a sequence of integers a1, a2, ..., an and q pairs of integers (l1, r1), (l2, r2), ..., (lq, rq), find count(l1, r1), count(l2, r2), ..., count(lq, rq) where count(i, j) is the number of different integers among a1, a2, ..., ai, aj, aj + 1, ..., an.
输入描述:
The input consists of several test cases and is terminated by end-of-file.The first line of each test cases contains two integers n and q.The second line contains n integers a1, a2, ..., an.The i-th of the following q lines contains two integers li and ri.
输出描述:
For each test case, print q integers which denote the result.
示例1
输入
复制
3 2
1 2 1
1 2
1 3
4 1
1 2 3 4
1 3
输出
复制
2
1
3
备注:
* 1 ≤ n, q ≤ 105* 1 ≤ ai ≤ n* 1 ≤ li, ri ≤ n* The number of test cases does not exceed 10.
这是唯一一题签到题 ，现场找的板子，修改一下过的
最简单粗暴的方法，把数组扩张一倍
修改一下查询区间。板子题的变形吧
1 #include <cstdio> 2 #include <iostream> 3 #include <algorithm> 4 #include <cmath> 5 #include <cstdlib> 6 #include <cstring> 7 #include <vector> 8 #include <list> 9 #include <map> 10 #include <stack> 11 #include <queue> 12 using namespace std; 13 #define ll long long 14 const int maxn1 = 2000005; 15 int cur[maxn1]; 16 int then[maxn1]; 17 int ans[maxn1]; 18 int limit, n, m; 19 struct node { 20 int l, r, id; 21 } que[maxn1]; 22 bool cmp(node x, node y) { 23 if(x.l / limit == y.l / limit) return x.r < y.r; 24 return x.l / limit < y.l / limit; 25 } 26 void solve() { 27 int L, R, ans1; 28 L = R = 0; 29 ans1 = 0; 30 for(int i = 1; i <= m; i++) { 31 while(que[i].l > L) { 32 then[cur[L]]--; 33 if(then[cur[L]] == 0) 34 ans1--; 35 L++; 36 } 37 while(que[i].r < R) { 38 then[cur[R]]--; 39 if(then[cur[R]] == 0) 40 ans1--; 41 R--; 42 } 43 while(que[i].l < L) { 44 L--; 45 then[cur[L]]++; 46 if(then[cur[L]] == 1) 47 ans1++; 48 } 49 while(que[i].r > R) { 50 R++; 51 then[cur[R]]++; 52 if(then[cur[R]] == 1) 53 ans1++; 54 } 55 ans[que[i].id] = ans1; 56 } 57 for(int i = 1; i <= m; i++) 58 printf("%d ", ans[i]); 59 } 60 int main() { 61 while(scanf("%d", &n) != EOF) { 62 scanf("%d", &m); 63 memset(cur, 0, sizeof(cur)); 64 memset(then, 0, sizeof(then)); 65 memset(ans, 0, sizeof(ans)); 66 for(int i = 1; i <= n; i++) 67 scanf("%d", &cur[i]); 68 for (int i = n + 1 ; i <= 2 * n ; i++) 69 cur[i] = cur[i - n]; 70 for(int i = 1; i <= m; i++) { 71 scanf("%d%d", &que[i].l, &que[i].r); 72 int temp1 = que[i].l, temp2 = que[i].r; 73 que[i].l = temp2, que[i].r = temp1 + n; 74 que[i].id = i; 75 } 76 limit = (int)(sqrt(2 * n) + 0.5); 77 memset(then, 0, sizeof(then)); 78 sort(que + 1, que + 1 + m, cmp); 79 solve(); 80 } 81 return 0; 82 }