题目链接：http://codeforces.com/problemset/problem/940/F
题目：
题意：求次数的mex，mex的含义为某个集合（如{1，2，4，5}）第一个为出现的非负数（3），注意是次数，而不是某个元素的mex。
思路：这一题数据太大，所以我们首先得进行一次离散化。用一个num2来记录每个次数出现次数，num1来记录次数出现次数，最后用一个for循环来求出mex。
代码实现如下：
1 #include <set> 2 #include <map> 3 #include <queue> 4 #include <stack> 5 #include <cmath> 6 #include <bitset> 7 #include <cstdio> 8 #include <string> 9 #include <vector> 10 #include <cstdlib> 11 #include <cstring> 12 #include <iostream> 13 #include <algorithm> 14 using namespace std; 15 16 typedef long long ll; 17 typedef unsigned long long ull; 18 19 #define bug printf("********* "); 20 #define FIN freopen("in.txt", "r", stdin); 21 #define debug(x) cout<<"["<<x<<"]" <<endl; 22 #define IO ios::sync_with_stdio(false),cin.tie(0); 23 24 const double eps = 1e-8; 25 const int mod = 1e9 + 7; 26 const int maxn = 1e5 + 7; 27 const double pi = acos(-1); 28 const int inf = 0x3f3f3f3f; 29 const ll INF = 0x3f3f3f3f3f3f3f3f; 30 31 inline int read() {//读入挂 32 int ret = 0, c, f = 1; 33 for(c = getchar(); !(isdigit(c) || c == '-'); c = getchar()); 34 if(c == '-') f = -1, c = getchar(); 35 for(; isdigit(c); c = getchar()) ret = ret * 10 + c - '0'; 36 if(f < 0) ret = -ret; 37 return ret; 38 } 39 40 int n, q, block, idq, idc, x, y; 41 int a[maxn], num1[2 * maxn], num2[2 * maxn]; 42 vector<int> v; 43 44 struct query { 45 int l, r, id, t, ans; 46 bool operator < (const query& x) const { 47 if((l - 1) / block != (x.l - 1) / block) { 48 return l < x.l; 49 } 50 if((r - 1) / block != (x.r - 1) / block) { 51 return r < x.r; 52 } 53 return t < x.t; 54 } 55 }ask[maxn]; 56 57 struct modify { 58 int p, pre, val; 59 }myf[maxn]; 60 61 int get_id(int x) { 62 return lower_bound(v.begin(), v.end(), x) - v.begin() + 1; 63 } 64 65 void add(int x) { 66 num1[num2[x]]--; 67 num2[x]++; 68 num1[num2[x]]++; 69 } 70 71 void del(int x) { 72 num1[num2[x]]--; 73 num2[x]--; 74 num1[num2[x]]++; 75 } 76 77 int main() { 78 //FIN; 79 num1[0] = 1e8; 80 n = read(); 81 q = read(); 82 block = 2000; 83 for(int i = 1; i <= n; i++) { 84 a[i] = read(); 85 v.push_back(a[i]); 86 } 87 int nw = 0; 88 for(int i = 1; i <= q; i++) { 89 int op; 90 op = read(); 91 if(op == 1) { 92 x = read(); 93 y = read(); 94 idq++; 95 ask[idq].l = x, ask[idq].r = y; 96 ask[idq].id = idq; 97 ask[idq].t = nw; 98 } else { 99 x = read(); 100 y = read(); 101 idc++; 102 nw++; 103 myf[idc].p = x; 104 myf[idc].pre = a[x]; 105 myf[idc].val = y; 106 a[x] = y; 107 v.push_back(y); 108 } 109 } 110 sort(v.begin(), v.end()); 111 v.erase(unique(v.begin(), v.end()), v.end()); 112 sort(ask + 1, ask + idq + 1); 113 for(int i = 1; i <= n; i++) { 114 a[i] = get_id(a[i]); 115 } 116 for(int i = 1; i <= idc; i++) { 117 myf[i].pre = get_id(myf[i].pre); 118 myf[i].val = get_id(myf[i].val); 119 } 120 int tmp = nw, r = 0, l = 1; 121 for(int i = 1; i <= idq; i++) { 122 int res = 1; 123 while(r > ask[i].r) { 124 del(a[r--]); 125 } 126 while(r < ask[i].r) { 127 add(a[++r]); 128 } 129 while(l > ask[i].l) { 130 add(a[--l]); 131 } 132 while(l < ask[i].l) { 133 del(a[l++]); 134 } 135 while(tmp < ask[i].t) { 136 tmp++; 137 if(myf[tmp].p >= l && myf[tmp].p <= r) { 138 del(myf[tmp].pre); 139 add(myf[tmp].val); 140 } 141 a[myf[tmp].p] = myf[tmp].val; 142 } 143 while(tmp > ask[i].t) { 144 if(myf[tmp].p >= l && myf[tmp].p <= r) { 145 del(myf[tmp].val); 146 add(myf[tmp].pre); 147 } 148 a[myf[tmp].p] = myf[tmp].pre; 149 tmp--; 150 } 151 while(num1[res] > 0) res++; 152 ask[ask[i].id].ans = res; 153 } 154 for(int i = 1; i <= idq; i++) { 155 printf("%d ", ask[i].ans); 156 } 157 return 0; 158 }