HDU 5183 Negative and Positive (NP) (set ＋ 读入外挂 乱搞)

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.

HintIf 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");
    }
}