BC31pairs&&hdoj5178&&hdoj5179

pairs

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1920    Accepted Submission(s): 681


Problem Description

John has n points
on the X axis, and their coordinates are (x[i],0),(i=0,1,2,…,n−1).
He wants to know how many pairs<a,b> that |x−x[a]|≤k.(a<b)

 

Input

The first line contains a single integer T (about
5), indicating the number of cases.

Each test case begins with two integers n,k(1≤n≤100000,1≤k≤109).

Next n lines
contain an integer x[i](−109≤x[i]≤109),
means the X coordinates.

 

Output

For each case, output an integer means how many pairs<a,b> that |x[b]−x[a]|≤k.

 

Sample Input

2
5 5
-100
0
100
101
102
5 300
-100
0
100
101
102

 

Sample Output

3
10

 

Source

BestCoder Round #31

 

注意要求的是abs(x[b]-x[a])<=k的个数

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
using namespace std;
const int maxn=100010;
int num[maxn];
int main()
{
int t,i,j,k,n;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&k);
for(i=0;i<n;++i){
scanf("%d",&num[i]);
}
sort(num,num+n);
int pos=0;
long long ans=0;
for(i=0;i<n;++i){
while(num[i]-num[pos]>k)pos++;
ans=ans+i-pos;
}
printf("%lld\n",ans);
}
return 0;
}

beautiful number

[b]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 509    Accepted Submission(s): 309


Problem Description

Let A=∑ni=1ai∗10n−i(1≤ai≤9)(n is
the number of A's
digits). We call A as
“beautiful number” if and only if a[i]≥a[i+1] when 1≤i<n and a[i] mod a[j]=0 when 1≤i≤n,i<j≤n(Such
as 931 is a "beautiful number" while 87 isn't).

Could you tell me the number of “beautiful number” in the interval [L,R](including
L and R)?

 

Input

The fist line contains a single integer T(about
100), indicating the number of cases.

Each test case begins with two integers L,R(1≤L≤R≤109).

 

Output

For each case, output an integer means the number of “beautiful number”.

 

Sample Input

2
1 11
999999993 999999999

 

Sample Output

10
2

 

Source

BestCoder Round #31

找出所有满足条件的数然后逐个判断即可

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
using namespace std;
const int maxn=2010;
int array[maxn],cnt=0;
void dfs(int d,int bit,int num){
array[cnt++]=num;if(d>=1000000000)return ;
for(int i=bit;i<=9;i+=bit){
dfs(d*10,i,num+d*i);
}
}
int main()
{
cnt=0;
for(int i=1;i<=9;++i){
dfs(10,i,i);
}
int i,j,k,t;
scanf("%d",&t);
while(t--){
int l,r,ans=0;
scanf("%d%d",&l,&r);
for(i=0;i<cnt;++i){
if(array[i]>=l&&array[i]<=r)ans++;
}
printf("%d\n",ans);
}
return 0;
}