hdu5183 hush

http://acm.hdu.edu.cn/showproblem.php?pid=5183

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.
 

/**hdu 5183  hash题目大意： NP?sum(i,j)=ai?ai+1+ai+2+?+(?1)j?iaj，求给定序列中是否有一个子序列得数为k解题思路：题目数据很大，只能用O(n)的复杂度。看了杭电的解题报告后，很久才理解。我求将序列的后缀和表示出来(sum[0]-sum[1]+sum[2]-……)          然后从后往前枚举所求区间的第一个位置i，若为奇数那么在h1的哈希表中寻找是不是存在sum[i]-k,如果没有将sum[i]插入哈希表。          如果i为偶数那么在h2中寻找-sum[i]-k，将-sum[i]插入哈希表中。          由于哈希表的插入和查寻复杂度接近O(1)，因此我们的算法复杂度总体为O(n)*/#include <stdio.h>#include <string.h>#include <algorithm>#include <iostream>using namespace std;#define LL long longconst int MAXN=1000010;const int HASH=1000007;struct HASHMAP{    int head[HASH],next[MAXN],ip;    LL state[MAXN];    void init()    {        memset(head,-1,sizeof(head));        ip=0;    }    bool check(LL val)    {        int h=(val%HASH+HASH)%HASH;        for(int i=head[h]; i!=-1; i=next[i])            if(state[i]==val)return true;        return false;    }    int insert(LL val)    {        int h=(val%HASH+HASH)%HASH;        for(int i=head[h]; i!=-1; i=next[i])        {            if(val==state[i])                return 1;        }        state[ip]=val,next[ip]=head[h],head[h]=ip++;        return 0;    }} h1,h2;LL a[MAXN];int n,k;int main(){    int T,tt=0;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&k);        for(int i=0; i<n; i++)            scanf("%I64d",&a[i]);        bool flag=false;        LL sum=0;        h1.init(),h2.init();        h1.insert(0),h2.insert(0);        for(int i=n-1; i>=0; i--)        {            if(flag==true)break;            if(i&1)                sum-=a[i];            else                sum+=a[i];            if(i%2==0)            {                if(h1.check(sum-k))                    flag=true;            }            else            {                if(h2.check(-sum-k))                    flag=true;            }            h1.insert(sum);            h2.insert(-sum);        }        printf("Case #%d: ",++tt);        if(flag)printf("Yes.\n");        else printf("No.\n");    }    return 0;}