BestCoder Round #36 (hdu5199)Gunner(水题)
转载请注明出处: http://www.cnblogs.com/fraud/ ——by fraud
Gunner
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Jack will shot many times, he wants to know how many birds fall during each shot.
a bullet can hit many birds, as long as they stand on the top of the tree with height of H.
In the second line, there are n numbers h[1],h[2],h[3],…,h[n] which describes the height of the trees.
In the third line, there are m numbers q[1],q[2],q[3],…,q[m] which describes the height of the Jack’s shots.
Please process to the end of file.
[Technical Specification]
1≤n,m≤1000000(106)
1≤h[i],q[i]≤1000000000(109)
All inputs are integers.
1 2 3 4
1 1 4
0
1
Huge input, fast IO is recommended.
水题
//#####################
//Author:fraud
//Blog: http://www.cnblogs.com/fraud/
//#####################
#include <iostream>
#include <sstream>
#include <ios>
#include <iomanip>
#include <functional>
#include <algorithm>
#include <vector>
#include <string>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <cctype>
using namespace std;
#define XINF INT_MAX
#define INF 0x3FFFFFFF
#define MP(X,Y) make_pair(X,Y)
#define PB(X) push_back(X)
#define REP(X,N) for(int X=0;X<N;X++)
#define REP2(X,L,R) for(int X=L;X<=R;X++)
#define DEP(X,R,L) for(int X=R;X>=L;X--)
#define CLR(A,X) memset(A,X,sizeof(A))
#define IT iterator
typedef long long ll;
typedef pair<int,int> PII;
typedef vector<PII> VII;
typedef vector<int> VI; int a[];
int b[];
int h[];
int p[];
int ans[];
bool cmp(int x,int y){
if(h[x]==h[y])return x<y;
return h[x]<h[y];
}
int Scan()
{
int res, ch=;
while(!(ch>=''&&ch<='')) ch=getchar();
res=ch-'';
while((ch=getchar())>=''&&ch<='')
res=res*+ch-'';
return res;
}
void Out(int a)
{
if(a>)
Out(a/);
putchar(a%+'');
} int main()
{
ios::sync_with_stdio(false);
int n,m;
while(scanf("%d%d",&n,&m)!=EOF){
for(int i=;i<n;i++)
a[i]=Scan();
sort(a,a+n);
b[]++;
int tot=;
for(int i=;i<n;i++){
if(a[i]==a[i-])b[tot]++;
else b[++tot]=;
}
for(int i=;i<m;i++)h[i]=Scan();
for(int i=;i<m;i++)p[i]=i;
sort(p,p+m,cmp);
int j=;
int k=;
for(int i=;i<m;i++){
if(j>=n)break;
while(a[j]<h[p[i]]){
j++;
if(j>=n)break;
if(a[j]!=a[j-])k++;
}
if(j>=n)break;
if(h[p[i]]==a[j]){
ans[p[i]]=b[k];
b[k]=;
}
}
for(int i=;i<m;i++){
Out(ans[i]);
puts("");
}
} return ;
}
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