hdu 5183 Negative and Positive (NP)（hash，思路）

Negative and Positive (NP)

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3330    Accepted Submission(s): 896

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

21 112 1-1 0

 

Sample Output

Case #1: Yes.Case #2: No.
Hint
If input is huge, fast IO method is recommended.

题意：给n个数字，问存不存在一个

NP−sum(i,j)=ai−ai+1+ai+2+⋯+(−1)j−iaj等于k

思路：我们先维护两个前缀和，第一个是sum[i]=a[1]-a[2]+a[3]+,,,+(-1)的(i-1)次方*a[i]

第二个就等于-sum[i]，就是说奇数项为负~

然后我们维护两个hash，一个保存sum[i],一个保存-sum[i]

然后枚举区间的左端点，因为sum[i]-sum[j]=k

因为区间[i,j]里，第一项a[i]必须为正。

所以如果当前枚举的i点下标为奇数，那么我们就从第一个hash里找有没有一个值等于sum[i]-k

反之如果当前枚举的i点下标为偶数，那么我们就从第二个hash里找有没有一个值等于-sum[i]-k(也就是把整个-sum[i]-k=-sum[j])

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;#define N 1000050long long a[N],sum[N];struct Hashmap{    int head[N],next[N],cnt;    long long state[N];    void init()    {        memset(head,-1,sizeof(head));        cnt=0;    }    int check(long long val)    {        int h=(val%N+N)%N;        for(int i=head[h];i!=-1;i=next[i])            if(state[i]==val) return 1;            return 0;    }    int Insert(long long val)    {        int h=(val%N+N)%N;        for(int i=head[h];i!=-1;i=next[i])            if(val==state[i])                return 1;        state[cnt]=val;        next[cnt]=head[h];        head[h]=cnt++;        return 0;    }}H1,H2;int main(){    int T;    int n,k,t=1;    scanf("%d",&T);    while(T--)    {        scanf("%d %d",&n,&k);        for(int i=1;i<=n;i++)            scanf("%lld",&a[i]);            long long sum=0;        H1.init();H2.init();        H1.Insert(0);H2.Insert(0);        int flag=0;        for(int i=n;i>=1&&!flag;i--)        {            if(i&1) sum+=a[i];            else sum-=a[i];            if(i&1)            {                if(H1.check(sum-k))                    flag=1;            }            else            {                if(H2.check(-sum-k))                    flag=1;            }            H1.Insert(sum);            H2.Insert(-sum);        }        printf("Case #%d: ",t++);        if(flag) printf("Yes.\n");        else printf("No.\n");    }    return 0;}

NP−sum(i,j)=ai−ai+1+ai+2+⋯+(−1)j−iaj