1002 Positive and Negative

题目链接

Negative and Positive (NP)

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1161    Accepted Submission(s): 59

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. 
    
   
   
   
   

/*
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。
总复杂度比o(n)大一点
*/

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
ll sum[1000000+5];
int T,n,k,x;
bool success;
set<ll>es;

/////////////////////G++条件下快速输入输出模板
template<class T>
inline bool scan_d(T &ret) {
    char c;
    bool sgn;
    if(c=getchar(),c==EOF)return 0;
    while(c!='-'&&(c<'0'||c>'9'))c=getchar();
    sgn=(c=='-');
    ret=sgn?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9')ret=ret*10+(c-'0');
    if(sgn)ret=-ret;
    return 1;
}
////////////////////////////sum[i]-sum[j]=k  ->   sum[i]-k=sum[j],进行枚举


int main() {
    //freopen("D://imput.txt","r",stdin);
    scanf("%d",&T);
    for(int i=1; i<=T; i++) {
        success=false;
        es.clear();
        scanf("%d%d",&n,&k);
        sum[0]=0;
        for(int j=1; j<=n; j++) {
            scan_d(x);
            if(j&1) {
                sum[j]=sum[j-1]+x;
            } else {
                sum[j]=sum[j-1]-x;
            }
        }
        for(int j=n; j>0; j--) {
            es.insert(sum[j]);
            if(j&1) {//奇偶不同导致他们的开头数字的正负不同，之后的数字和则会相反
                if((j&1) && es.find(sum[j-1]+k)!=es.end()||!(j&1) && es.find(sum[j-1]-k)!=es.end()) {
                    success=true;
                    break;
                }
            }
        }
        printf("Case #%d: ",i);
        if(success) {
            printf("Yes.\n");
        } else {
            printf("No.\n");
        }
    }
    return 0;
}