题目链接:http://codeforces.com/contest/137/problem/B
给n个数字,要求修改成1~n的全排列数中的一个,修改的次数尽可能少,问最少需要修改几个数。
记下数组里出现的数字都是什么,之后处理的时候,遇到出现大于1次的说明要修改。有一个trick那就是ai<=5000,也就是会超出n,所以要特别注意一下,在这里WA了一次。
1 /* 2 ━━━━━┒ギリギリ♂ eye! 3 ┓┏┓┏┓┃キリキリ♂ mind! 4 ┛┗┛┗┛┃\○/ 5 ┓┏┓┏┓┃ / 6 ┛┗┛┗┛┃ノ) 7 ┓┏┓┏┓┃ 8 ┛┗┛┗┛┃ 9 ┓┏┓┏┓┃ 10 ┛┗┛┗┛┃ 11 ┓┏┓┏┓┃ 12 ┛┗┛┗┛┃ 13 ┓┏┓┏┓┃ 14 ┃┃┃┃┃┃ 15 ┻┻┻┻┻┻ 16 */ 17 #include <algorithm> 18 #include <iostream> 19 #include <iomanip> 20 #include <cstring> 21 #include <climits> 22 #include <complex> 23 #include <fstream> 24 #include <cassert> 25 #include <cstdio> 26 #include <bitset> 27 #include <vector> 28 #include <deque> 29 #include <queue> 30 #include <stack> 31 #include <ctime> 32 #include <set> 33 #include <map> 34 #include <cmath> 35 using namespace std; 36 #define fr first 37 #define sc second 38 #define cl clear 39 #define BUG puts("here!!!") 40 #define W(a) while(a--) 41 #define pb(a) push_back(a) 42 #define Rint(a) scanf("%d", &a) 43 #define Rll(a) scanf("%I64d", &a) 44 #define Rs(a) scanf("%s", a) 45 #define Cin(a) cin >> a 46 #define FRead() freopen("in", "r", stdin) 47 #define FWrite() freopen("out", "w", stdout) 48 #define Rep(i, len) for(int i = 0; i < (len); i++) 49 #define For(i, a, len) for(int i = (a); i < (len); i++) 50 #define Cls(a) memset((a), 0, sizeof(a)) 51 #define Clr(a, x) memset((a), (x), sizeof(a)) 52 #define Fuint(a) memset((a), 0x7f7f, sizeof(a)) 53 #define lrt rt << 1 54 #define rrt rt << 1 | 1 55 #define pi 3.14159265359 56 #define RT return 57 #define lowbit(x) x & (-x) 58 #define onenum(x) __builtin_popcount(x) 59 typedef long long LL; 60 typedef long double LD; 61 typedef unsigned long long Uint; 62 typedef pair<int, int> pii; 63 typedef pair<string, int> psi; 64 typedef map<string, int> msi; 65 typedef vector<int> vi; 66 typedef vector<int> vl; 67 typedef vector<vl> vvl; 68 typedef vector<bool> vb; 69 70 const int maxn = 5050; 71 int n; 72 int a[maxn]; 73 int vis[maxn]; 74 75 int main() { 76 // FRead(); 77 while(~Rint(n)) { 78 Cls(vis); 79 For(i, 1, n+1) { 80 Rint(a[i]); 81 vis[a[i]]++; 82 } 83 int ret = 0; 84 For(i, 1, n+1) { 85 if(a[i] <= n) { 86 if(vis[a[i]] > 1) { 87 vis[a[i]]--; 88 ret++; 89 } 90 } 91 else { 92 vis[a[i]]--; 93 ret++; 94 } 95 } 96 printf("%d ", ret); 97 } 98 RT 0; 99 }