Negative and Positive (NP)
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2846 Accepted Submission(s): 782
Problem Description
When given an array (a0,a1,a2,?an?1) and an integer K, you are expected to judge whether there is a pair (i,j)(0≤i≤j<n) which makes that NP?sum(i,j) equals to K true. Here NP?sum(i,j)=ai?ai+1+ai+2+?+(?1)j?iaj
Input
Multi test cases. In the first line of the input file there is an integer T indicates the number of test cases.
In the next 2?T lines, it will list the data for each test case.
Each case occupies two lines, the first line contain two integers n and K which are mentioned above.
The second line contain (a0,a1,a2,?an?1)separated by exact one space.
[Technical Specification]
All input items are integers.
0<T≤25,1≤n≤1000000,?1000000000≤ai≤1000000000,?1000000000≤K≤1000000000
Output
For each case，the output should occupies exactly one line. The output format is Case #id: ans, here id is the data number starting from 1; ans is “Yes.” or “No.” (without quote) according to whether you can find (i,j) which makes PN?sum(i,j) equals to K.
See the sample for more details.
Sample Input
2
1 1
1
2 1
-1 0
Sample Output
Case #1: Yes.
Case #2: No.
Hint
If input is huge, fast IO method is recommended.
Source
BestCoder Round #32
Positive and Negative
维护前缀和sum[i]=a[0]-a[1]+a[2]-a[3]+…+(-1)^i*a[i]
枚举结尾i，然后在hash表中查询是否存在sum[i]-K的值。
如果当前i为奇数，则将sum[i]插入到hash表中。
上面考虑的是从i为偶数为开头的情况。
然后再考虑以奇数开头的情况，按照上述方法再做一次即可。
不同的是这次要维护的前缀和是sum[i]=-（a[0]-a[1]+a[2]-a[3]+…+(-1)^i*a[i]）
I为偶数的时候将sum[i]插入到hash表。
总复杂度o(n)
#include<bits/stdc++.h>
using namespace std;
template<class T>inline T read(T&x)
{
char c;
while((c=getchar())<=32)if(c==EOF)return 0;
bool ok=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=1e6+1000;
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 100007
LL sum[maxn];
LL a[maxn];
vector<LL> hash_[mod];
void insert(LL key)
{
int pos=(key%mod+mod)%mod;
hash_[pos].push_back(key);
}
bool find(LL key)
{
int pos=(key%mod+mod)%mod;
int N=hash_[pos].size();
for(int i=0;i<N;i++)
if(hash_[pos][i]==key)return true;
return false;
}
bool solve(LL n,LL k)
{
for(int i=0;i<mod;i++)hash_[i].clear();
insert(0);
for(int i=0;i<n;i++)
{
if(find(sum[i]-k))return true;
if(i&1)insert(sum[i]);
}
for(int i=0;i<mod;i++)hash_[i].clear();
for(int i=0;i<n;i++)
{
if(find(-sum[i]-k))return true;
if(!(i&1))insert(-sum[i]);
}
return false;
}
int main()
{
int T,cas=1;
read(T);
while(T--)
{
LL k,n;
read_(n,k);
for(int i=0;i<n;i++)
{
read(a[i]);
}
sum[0]=a[0];
for(int i=1;i<n;i++)
if(i&1)
sum[i]=sum[i-1]-a[i];
else
sum[i]=sum[i-1]+a[i];
printf("Case #%d: %s\n",cas++,solve(n,k)?"Yes.":"No.");
}
return 0;
}