HDU 5183 Negative and Positive (NP)(哈希 hash)——BestCoder Round #32
Negative and Positive (NP)
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
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
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.
See the sample for more details.
Sample Input
2 1 1 1 2 1 -1 0
Sample Output
Case #1: Yes. Case #2: No.HintIf input is huge, fast IO method is recommended.
Source
BestCoder Round #32
/************************************************************************/
附上该题对应的中文题
Negative and Positive (NP)
Time Limit: 3000/1500 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
问题描述
给定一个数组(a0,a1,a2,⋯an−1)和一个整数K， 请来判断一下是否存在二元组(i,j)(0≤i≤j<n)使得 NP−sum(i,j) 刚好为K。这里。
输入描述
多组测试数据。在文件的第一行给出一个T，表示有T组数据。 在接下来的2∗T行里，将会给出每一组数据。 每一组数据占两行，第一行包含n和K。 第二行包含(a0,a1,a2,⋯an−1)以一个空格分开。 [参数说明] 所有输入均为整数。 0<T≤25,1≤n≤1000000,−1000000000≤ai≤1000000000,−1000000000≤K≤1000000000
输出描述
对于每一个数据，输出占一行，输出格式是Case #id: ans,这儿id是数据编号，从1开始，ans是根据是否找到满足的二元组而定为“Yes.” 或 “No.” (不包含引号) 看样例可以获得更多的信息。
输入样例
2 1 1 1 2 1 -1 0
输出样例
Case #1: Yes. Case #2: No.
Hint
如果数据比较多，建议使用快速读入。
/****************************************************/
出题人的解题思路：
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)该题的解法，出题人讲得很明确，利用哈希函数，我们将求得的sum[i](前i项的NP-sum值)求出来放入哈希表中，遍历一遍即可判断是否存在一个二元组满足条件
我要提的一点是题目的读入优化，因为题目的数据比较多，普通的输入耗的时间也会比较长，容易TLE，所以此题需用到读入优化，详情可点击链接
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
#define ll __int64
using namespace std;
const int N = 1000005;
const int inf = 1000000000;
const int mod = 1000005;
struct node
{
int to;
ll v;
}e[N];
int s[N],h[N],p;
void add(ll x)
{
e[p].v=x;
x=abs(x)%N;
e[p].to=h[x];
h[x]=p++;
}
bool check(ll x)
{
int k=abs(x)%N;
for(int i=h[k];i+1;i=e[i].to)
if(e[i].v==x)
return true;
return false;
}
int get_val(int &ret)
{
int sgn=1;
char c;
while(((c=getchar())<'0'||c>'9')&&c!='-');
if(c=='-')
sgn=-1,ret=0;
else
ret=c-'0';
while((c=getchar())>='0'&&c<='9')
ret=ret*10+c-'0';
ret*=sgn;
}
int main()
{
int t,n,k,i,j=1;
ll sum;
get_val(t);
while(t--)
{
sum=0;p=0;
memset(h,-1,sizeof(h));
get_val(n);
get_val(k);
for(i=0;i<n;i++)
get_val(s[n-i-1]);
add(0);
printf("Case #%d: ",j++);
for(i=0;i<n;i++)
{
sum+=(i&1)?-s[i]:s[i];
add(sum);
if(i&1)
{
if(check(sum+k))
{
puts("Yes.");
break;
}
}
else if(check(sum-k))
{
puts("Yes.");
break;
}
}
if(i>=n)
puts("No.");
}
return 0;
}
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