Negative and Positive (NP)

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 502    Accepted Submission(s): 141

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

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

题目大意：问是否存在(i, j，使)NP?sum(i,j)=ai?ai+1+ai+2+?+(?1)j?iaj这个东西算出来的值等于k，存在输出Yes，否则输出No

题目分析：这题时限太死，水过的，Hint里面提示用fast IO也就是读入外挂了，主要做法是处理一下前缀和，第一种是正负正，第二种是负正负，对于要求的k，我们边插边找，找的时候要分奇偶，若i为奇数，则在第一种情况中找，否则在第二种情况中找，找到就退出

#include <cstdio>
#include <cstring>
#include <set>
#define ll long long
using namespace std;
int const MAX = 1e6 + 5;
ll a[MAX], sum[MAX];
set <ll> s;

ll Scan()
{
    ll res = 0, ch, flag = 0;
    if((ch = getchar()) == '-')
        flag = 1;
    else if(ch >= '0' && ch <= '9')
        res = ch - '0';
    while((ch = getchar()) >= '0' && ch <= '9' )
        res = res * 10 + ch - '0';
    return flag ? -res : res;
}  

int main()
{
    int T, n, k;
    scanf("%d", &T);
    for(int ca = 1; ca <= T; ca++)
    {
        bool flag = false;
        s.clear();
        memset(sum, 0, sizeof(sum));
        scanf("%d %d", &n, &k);
        getchar();
        for(int i = 1; i <= n; i++)
            a[i] = Scan();
        for(int i = 1; i <= n; i++)
            sum[i] = sum[i - 1] + (i % 2 ? a[i] : -a[i]);
        for(int i = n; i > 0; i--)
        {
            s.insert(sum[i]);
            if(i % 2)
            {
                if(s.find(sum[i - 1] + k) != s.end())
                {
                    flag = true;
                    break;
                }
            }
            else
            {
                if(s.find(sum[i - 1] - k) != s.end())
                {
                    flag = true;
                    break;
                }
            }
        }
        printf("Case #%d: %s.\n", ca, flag ? "Yes" : "No");
    }
}