codeforces1352 E. Special Elements
Pay attention to the non-standard memory limit in this problem.
In order to cut off efficient solutions from inefficient ones in this problem, the time limit is rather strict. Prefer to use compiled statically typed languages (e.g. C++). If you use Python, then submit solutions on PyPy. Try to write an efficient solution.
The array a=[a1,a2,…,an]
(1≤ai≤n) is given. Its element ai is called special if there exists a pair of indices l and r (1≤l<r≤n) such that ai=al+al+1+…+ar
. In other words, an element is called special if it can be represented as the sum of two or more consecutive elements of an array (no matter if they are special or not).
Print the number of special elements of the given array a
.
For example, if n=9
and a=[3,1,4,1,5,9,2,6,5], then the answer is 5
:
a3=4
is a special element, since a3=4=a1+a2=3+1
;
a5=5
is a special element, since a5=5=a2+a3=1+4
;
a6=9
is a special element, since a6=9=a1+a2+a3+a4=3+1+4+1
;
a8=6
is a special element, since a8=6=a2+a3+a4=1+4+1
;
a9=5
is a special element, since a9=5=a2+a3=1+4
.
Please note that some of the elements of the array a
may be equal — if several elements are equal and special, then all of them should be counted in the answer.
Input
The first line contains an integer t
(1≤t≤1000) — the number of test cases in the input. Then t
test cases follow.
Each test case is given in two lines. The first line contains an integer n
(1≤n≤8000) — the length of the array a. The second line contains integers a1,a2,…,an (1≤ai≤n
).
It is guaranteed that the sum of the values of n
for all test cases in the input does not exceed 8000
.
Output
Print t
numbers — the number of special elements for each of the given arrays.
Example
Input
Copy
5
9
3 1 4 1 5 9 2 6 5
3
1 1 2
5
1 1 1 1 1
8
8 7 6 5 4 3 2 1
1
1
Output
Copy
5
1
0
4
0
让你找出这个序列里的 “特殊数”的和：该数可以表示成连续子序列的和
WA代码：尺取的时候没有道理可言，造成有些子序列丢失，丢失答案
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<iostream>
using namespace std;
int flag[10000] ; //标记次数
int a[10000];
int main()
{
int t ;
scanf("%d",&t);
while(t--)
{
memset(flag,0,sizeof(flag));
int n;
int maxnum=-1;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
maxnum=max(maxnum,a[i]);
flag[a[i]]++;
}
int l=1,r=1;
int sum=a[l];
int ans=0;
while(l<n)
{
sum+=a[r+1]; //右指针向右试探
if(sum<=maxnum && r<n)
{
r++;
if(flag[sum] && r-l>0)
{
ans+=flag[sum];
flag[sum]=0;
}
}
else
{
if(l<r)
{
sum-=a[r+1]; //试探值复原
sum-=a[l];
l++; //左指针向右走
if(flag[sum] && r-l>0)
{
ans+=flag[sum];
flag[sum]=0;
}
}
else //l==r && 试探>maxnum
{
sum-=a[r+1]; //试探值复原
sum-=a[l];
l++;
r++;
sum+=a[r]; //此时序列就一个元素
}
}
// cout<<"ans: "<<ans<<endl;
// cout<<"l: "<<l<<" r: "<<r<<endl;
}
cout<<ans<<endl;
}
return 0;
}
比较笨的做法：前缀和
AC 代码：
#include<cstdio>
#include<cstring>
#include<iostream>
#include<string>
#include<algorithm>
#include<cmath>
#include<map>
using namespace std;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
int a[10000];
int sum[10000];
int flag[10000];
memset(flag,0,sizeof(flag));
memset(sum,0,sizeof(sum));
int maxnum=-1;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
maxnum=max(maxnum,a[i]);
sum[i]=a[i]+sum[i-1];
flag[a[i]]++;
}
int cnt=0;
for(int i=0;i<=n;i++)
{
for(int j=i+2;j<=n;j++)
{
if(sum[j]-sum[i]>maxnum)
break;
else
{
if(flag[ sum[j]-sum[i] ])
{
cnt+=flag[ sum[j]-sum[i] ];
flag[ sum[j]-sum[i] ]=0;
}
}
}
}
cout<<cnt<<endl;
}
return 0;
}
