Delete
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 207 Accepted Submission(s): 140
Problem Description
WLD likes playing with numbers. One day he is playing with N integers. He wants to delete K integers from them. He likes diversity, so he wants to keep the kinds of different integers as many as possible after the deletion. But he is busy pushing, can you help him?
Input
There are Multiple Cases. (At MOST 100)
For each case:
The first line contains one integer N(0 The second line contains N integers a1,a2,...,aN(1≤ai≤N), denoting the integers WLD plays with. The third line contains one integer K(0≤K Output For each case: Print one integer. It denotes the maximum of different numbers remain after the deletion. Sample Input 4 1 3 1 2 1 Sample Output 3 Hint if WLD deletes a 3, the numbers remain is [1,1,2],he'll get 2 different numbers. if WLD deletes a 2, the numbers remain is [1,1,3],he'll get 2 different numbers. if WLD deletes a 1, the numbers remain is [1,2,3],he'll get 3 different numbers. Source BestCoder Round #39 ($) 1001 Delete 用一个cnt数组记下每个数在a序列中出现了几次 在删数的时候贪心，尽可能删那些出现次数>1的数 这样就可以使最后有最多不同的数 #include<iostream>#include<cstring>#include<cstdio>#include#include<set>#include<map>#include<vector>#include<stack>#include<queue>#include<string>#include<cstdlib>#include<cmath>//#include<b>using namespace std;template<class t="">inline T read(T&x){ char c; while((c=getchar())<=32)if(c==EOF)return 0; bool k=false; if(c=='-')ok=true,c=getchar(); for(x=0; c>32; c=getchar()) x=x*10+c-'0'; if(ok)x=-x; return 1;}template<class t="">inline T read_(T&x,T&y){ return read(x)&&read(y);}template<class t="">inline T read__(T&x,T&y,T&z){ return read(x)&&read(y)&&read(z);}template<class t="">inline void write(T x){ if(x<0)putchar('-'),x=-x; if(x<10)putchar(x+'0'); else write(x/10),putchar(x%10+'0');}template<class t="">inline void writeln(T x){ write(x); putchar('\n');}//-------ZCC IO template------const int maxn=500;const double inf=999999999;#define lson (rt<<1),L,M#define rson (rt<<1|1),M+1,R#define M ((L+R)>>1)#define For(i,t,n) for(int i=(t);i<(n);i++)typedef long long LL;typedef double DB;typedef pair<int,int>P;#define bug printf("---\n");#define mod 100007int a[maxn];bool cmp(int a,int b){ return a>b;}int main(){ int n,m; while(read(n)) { memset(a,0,sizeof(a)); int maxv=0; For(i,0,n) { int tmp; read(tmp); a[tmp]++; maxv=max(maxv,tmp); } int k; read(k); sort(a,a+101,cmp); int i=0; while(a[i])i++; int sum=0; For(j,0,i)sum+=a[j]; sum-=i; k-=sum; if(k<=0)writeln(i); else writeln(i-k<0?0:i-k); } return 0;}</int,int></class></class></class></class></class></bits></cmath></cstdlib></string></queue></stack></vector></map></set></cstdio></cstring></iostream>